SECTION D: ELECTROSTATICS, CAPACITORS AND ELECTRICITY Electrostatics (static electricity) is the study of electric charges at rest, the forces between them and the clectric fields cssocioted with them. Static electricity occurs when positive (+) or negative (-) electrical charges collect on an object's surface. There are several methods through which this condition can be caused. (One way is by rubbing certain materials together or pulling them apart. Another way é by bringing a charged material near to c neutral material, and ao by sharing the charge on a body with mother neutral insulated body when they come into contact with each other, Electrification by friction / charging by rubbing or friction When two disimilar bodies are rubbed together, heat is generated due to friction The heat is sufficient to make the material of lower work function to release some electron, which are taken up by other material, The one which lost electrons become positively charged while the one which gained electrons becomes negatively charged > The number of electrons lost s equal to the number of electrons acquire therefore two insulating bodies rubbed together acquire equal and opposite charges. Examples of charging by friction = When a polythene rod (ebonite rod) is rubbed with fur (woolen duster), the ebonite rod becomes negatively charged while the duster becomes positively charged. = Ifa glass rod (cellulose acetate) is rubbed with silk, a glass rod becomes positively charged while the sik becomes negatively charged. vv 7 Insulators, semiconductors and conductors Conductor This is a material with free electrons and it con allow electricity and heat to pass through it. Exermpless Copper, bronze Insulator This b «material without free electrons and it cannot allow electricity and heat to pass through, it, Examples: Dry wood, plastic Semiconductors These are materials which allow electric charges to pass through them with difficulty. Examples: Moist air, paper Law of electrostatics Like charges repel each other and unlike charges attract each other. Precautions taken when carrying experiments in electrostatics @ Apparatus must be insulated The surrounding must be free from dust and moisture Attraction of neutral body by charged body Consider the uncharged conductor being brought near a negatively charged ebonite rod Negative charges on the ebonite rod repel the free electrons on the conductor to the remote ‘end and positive charge is thus left near the end of the metal adjacent to the ebonite rod. So the conductor & now attracted by the ebontte rod. 243GOLD LEAF ELETROSCOPE (GLE) Bass ae Uses of GLE (Test for the presence of charge ‘class window an (i) Test the sign of the charge tea i car (iff) To test the magnitude of charge erass pial} | aca (fv) Mecsure p.d jectrostatic induction It’s a phenomenon that describes the formation of charges on a conductor when a charged body is brought necr it. The charge acquired is opposite to that of inducing body. Charging a gold leat electroxope by induction (@) Charging G.L.E negatively Positively charged tod See Ee Brass cap te ‘Gold leat Brass plate “ \ * A positively charged glass rod is brought near the cap of the G.LE, negative charges are (&) Charging G.L.E negatively Negatively charged 710d Brash eap Bold leat | Brass plate + A negatively charged rod is brought near the cap of the G.LE, positive charges are induced induced on the brass cap and positive charges on GL amd brass plate. The gold leaf diverges. + With glass rod still in position, the G.LE is earthed, Free electrons flow from the earth to the brass plate and gold leaf thus collapses. ++ With the rod still in position, the earthing wire bremoved. + Glass rod is removed, the negative charges then redistribute thernselves to the brass cap, plate and gold leaf thus causing he leaf to diverge. The electroscope is now negatively charged. on the brass cap and negative charges on G.L and brass plate. The gold leaf diverges. + With glass rod still in position, the GLE is earthed, Free electrons flow from the plate and leaf to the earth thus the leaf collapses. + With the rod still in position, the earthing wire is removed. The rod is removed, the positive charges then redistribute therrselves to the brass cap, plate and gold leaf thus causing he leaf to diverge. ‘The electroscope is now positively charged Testing for the sign of charge on a body + Charge an electroscope negatively and the divergence noted. Bring the body under test near the cap of GLE Ifthe leaf divergence increases then that body is negatively charged, but if the divergence of the leaf decreases, then that body has either positive charge or it s neutral body + To differentiate between the two oitematives, discharge the GLEand now charge it positively + Bring the same body under test near the cap of appositively charged GLE If the leaf divergence increases again, then that body has positive charges but if the leaf divergence decreases then that body is neutrai. 244Notes Repubion és the only confirmatory test for the sign of the charge Summary - - Increasecepursion) - ‘Decrease (attraction) H i Decrease (attraction) ae Uncharaed boay Decrease (attraction) Charging a conductor by induction electrons in the metal sphere are repelled to the far end of the sphere. + The sphere is earthed while the charged body is still in position. Free electrons move from the sphere to the earth. + The earthing wire is removed while the charged tod is still in position + The charged body is removed and charges distributes therrselves all over the sphere. Hence the metal sphere is now positively charged. + Metal sphere on em insulating stand is placed near the negatively charged body. Free b) Negatively posinvely electrons in the metal sphere are attracted to » the near end of the sphere. | The spheres earthed while the charged body still in postion, Free electrons move from the earth to the sphere. | The earthing wire is removed while the charged rod i stillin position | The charged body is removed and charges distributes therreelves allover the sphere. Hence the metal sphere is now negatively stand + Metal sphere on an insulating stand is placed near the postively charged body. Free oes: Separation of conductors ) ea aia “Two identical brass spheres A and B are placed Z together so that they touch one another. fi) 245,ALB A positively charged rod is now brought near G end A and as arresuit negative charge is induced at A and postive charges repelled to B Keeping the positive rod in position, sphere B is moved a short distance away from B iv) aie The chenged rod is now removed and. S chongu reanrbute Explain how two spherical conductors made of brass can be changed oppositely and simultaneously by induction. How to distinguish a conductor and an insulator using an electroscope An electroscope i given charge and the divergence noted. The material s brought near the cap of the electroscope + If there is no change in divergence, the material isan insulator. If the leaf divergences material b « conductor Charging a bedy negatively at zero potential A posttively charged glass rod is brought near end A of the conductor. Negative charges are induced at the near end and positive charges at the far end of the neutral body. +} With the glass rod stillin position, body bs earthed. Body is now negatively charged at zero potential lectrophorus ‘This an instrument for produce unlimited supply of charge but it snot source of energy though converts mechanical energy to electrical energy Distribution of charge over the surface of a conductor, surface charge density: This s the quantity of charge per unit area over the surface of the conductor. Charge is mostly concentrated at sharp points <> ee Note: Charge only resides on the outside of a hollow conductor Investigating charge distribution on a pear shaped conductor 246+ Approof plane is placed on the surface of the conductor. A sample of charge acquired by the proof plane é then transferred to a hollow metal can placed on the cap neutral electroscope and the deflection of the electroscope és noted + The proof plane is then used to pick samples charges from different parts of the conductor ond each time the deflection of the electroscope s noted + The greatest deflection is obtained when the sample of charge are picked from the pointed end of the conductor. Therefore the surface charge density of charges 6 greatest where the curvature is areatest Experiment to show distribution of charge in a hollow conductor. (Faraday’s ice pail experiment) Inuling tread _— + Sis withdrawm, the leaf of the electroscope collapses y IK + Sis again lowered inside the metal ( without JIN touching i), the leaf of the electroscope / : diverges to the same extent es before. Neal | J . + Sisthen allowed to touch the can. The (t)s | inuaw divergence of the leaf remains unchanged Wy yy Sis withdrawn and on testing, it i found to ATT OTT have no cherge 4 There must have been charge inside the can scuatand opp tothe charge on .rince ee aunt beeing otha oul eof Th connected to a gol lectroscope. The ° ‘ sai anal lsabbibilotenaue cherae equal to that which we rial on + A posttively charged metal sphere, 5 is Action at sharp points [Corona discharge] The high electric field intensity at the sharp points of a charged conductor, ionizes the air molecules around the sharp points. The ions of opposite charge are attracted to the sharp point and neutralize the charge there, This way the conductor loses charge and the process is called corena discharge. Applications of action at sharp points (@) Lightening Conductor Action of alistening conducter SSS ety sages ot > When a charged cloud passes over a alittle lightening conductor, it induces opposite BS ® oomacraue charge on the spikes of the conductor which Seales results to high electric field intensity sone > The high electric field intensity on the spikes fonizes the air cround it. Charge similar to those on the spikes i repelled to the clouds ‘and neutralize charge on the cloud, while connect those opposite are attracted and discharged at the spikes cooper to > This way charge from the cloud is safely conducted to the ground Effect of Lightening Cloudk in relative motion become charged due to friction. The resulting charge builds up leading to, high p.d between the clouds and the earth. Large discharge currents through the building can cause themto bum 247(b) Vander sract generator ee ~ round it repelling positive charges on to the belt. ‘The belt driven by a motor carries this charge into the sphere. As it approaches F,, it induces negative charge at the spikes of E, and positive charge on the sphere Setaeracs “ The high electric field intensity around E, ionizes air there, repelling negative charge onto the belt. The negative charge neutralizes positive charge on the belt before it goes over the upper pulley. + The process is repeated many times until the oP The electrode £, is meade 10°V positive potential of the sphere is about 10°V relative relative to the earth. The high electric field tothe earth intensity at the sharp points of £, ionizes air COULOMBS LAW OF ELECTROSTATICS It states that the force between any two point charges b directly proportional to the product of the ‘magnitude of the charges emd inversely proportional to the square of the distance of separation of the charges, po ake * Gre? For vacuum or air ~ 0: Freyr? & = 885x10- 7 Fm-* = 9x10° ey r ae F This law holds for alllsign of charges. If Q, and Q; are unlike then the fore. like then the force is repubive Example Mustration Find the force between two point charges +40 and —3uC placed 10cm apart Solution attractive but if they are Attractive reine © | resi pa 2210 | 9x10°x4x1076x3x107% | F = 10.8N attractive ana ttE FTO aa 2 Three point charges are placed at point A, B and C as shown below A Sem “sue +anc os Find the resultemt force on the charge at B Solution 248ma p= 2x1 adtO-E7 10°F $: 3 —2 0.05 2102 = 100.8 attracted towards C eerie ESTEE Fy = 1008-108 9x10" x4x107 x32 10" Fy = 90N attracted towards C ole 10.8 attracted towards A 3. 3 dom ane ne ee Cakulate the force on —3p:0 Solution 5 acm pp = De adt0-FBx10-% 1 rae nee tien aR a Re. = oF ane? F. = 67.5N towards left py = setae se10"+ Fy = 1354675 ae ana? Fy = 202.5N towenrds let F, = 135N towards left a om. 6m ioe ae ? Find the resultant force acting at point P if a charge of 11 is placed at point P. Solution Stat ay 9xl0%x2x10~%x 1x10" SF ae R= e = 2:0 at = 2% F, = 0.05N towards right i ‘e10%xt10Exa.030"° Fp = 0.05 ~ 0.027 ee aaane aa Fp = 0.023 towards right Fy = 0.027N towards left 5 Find the resultemt force at C Solution FiaMonesen tan Fy, = 33.75N S10 2ex10"txS410% fs 0a Fy = 135Nupwards Px10~x3x10* f 04? Fr = 84.375Ntowends right P= (Saya) * (igs) 09") 60.57 F=(4rn) 249F? = 6057 + 111.17? F = 12659N a= tan" ¢ to) = 61.42" ~ e057) ° Resultant forces at Ck 2659N at 61.42" tothe horizontal Fy =2.7N 37Scos758) , ( 2:7c05755 33] 56in75.8) * \—2.7sin75. 7 (o6ss) Fs A P= 1521? +0653 a 1.655NV 9x10" xSx 107x321 07° si O27 tana = T5321 pean a=2333° Rrieecaurecsee Resuitemt forces et C & L6SSN ot 23.23° to the Fi oF horizontal 7. Four point charges Q:,(2,(s and Q,are placed at different comets of rectangle wo uc Qu Qs om Qs a, Cae -3ac Find the resultant force at Q» Solution 2uc 0 =53.13° 210s neor? Suc _ omer te 008 ‘ F, = SON doumwends Suro? asrt0rexae10° 4.4 N towards the right fF; = Seleexsat0~xze0°8 4 0.05 250F = 36N meee st) r= (88 tana 71. 55.6 em a =52.2° i Resultant forces at Qs s 90.66N at 52.2° tothe horizontal 8. Two points charges A nad B of —17.6xC and —9.0y:C respectively are placed in vacuum at a distance of 2m apart. When atthird charge C is placed mid way between A and B, the net force on B is zero @ Determine the charge on C Gi) Sketch the electric field lines for the above charge distribution Solution py = atenzantertonsere Fo Fe Qx10°x17.6x10°Fx9x10°% _ 9x 10°x9x10-FxQ 021 0.1057 Q= 44x10-°C % Two pith balls Pand Q each of mass 0.1g are separately suspended form the same point by threads 30cm long. When the balks are given equal charges, they repel each other and come to rest 18cm apert. Find the magnitude of each charge. 9=175° (1) Toso = mg Teast7.5° = 0.1x10°x9.81 T= 1.029x10"'N @ Cyrene = sonert0~ty 2 = 2102? e350 ~ (0.18) Q=3.33x10°C sind =, 10. Two identical conducting balls of mass m are each suspended in ai from a thick of length |, When the two balls are each given identical charge Q, they move apart as shown below at equilibrium each thread makes an angle 0 with vertical and separated x is given by a \%s (eens) 2512 Py Q Gre,mgtand But for small angles in radians tand = sin x 2 Exercise 1 Two points charges A nad B of 47.00 and 24.0uC respectively are placed in vacuum at a distance of 30cm apart. When aithird charge C of —35.0uC is placed between A and B at a distance of 20cm from A. find the net force on C An(-643.2N) 2. Two point charges of 5110 and 2:C are placed in liquid of relative permittivity 9 at distance Sem apart. Cakulate the force between them An(3.996N) 3+ Two insulating metal spheres each of charge 510°°C are separated by distance of 6cm. What is the force of repubion it; (@) The spheres are in air An(0.00625N) () The spheres are in air with the charge in each sphere doubled and their distance apart is halved An(O.1N) © The two sphere are placed in water whose dielectric constant is 81 Arm(7.7x10-°N) a 4.5 HC ‘Sem Bu Fem 25uc Find the resultant force at charge B. An(12.58N at 88.5 °to the horizontal ) 5. Aoi “SC Cakulate the resultant force at charge O, where O is the mid- point of the square An(523. 166N at 46.8°to the horizontal) 252ELECTRIC FIELD An electric field is a region within which on electric force is experienced. Electric fields cam be represented by electric field line. Definition ‘An electric field line is the path taken by a smal positive charge placed in the field Properties of electric field lines They otiginate from positive and end on negative. they are in a state of tension which causes them to shorten they repel one another side ways they travel in straight lines and never cross each other the number of fied lines originating or terminating on a charge is proportional to the magnitude of the charge Field patterns f) —_Iolated positive charge fi) olated negative charge fii) Two equal opporite charges 'v) Two poritive charges near each other vi) Two negative charges near each other X ig Neutral pointy Neutral point is the point in the electric field where the resultamt electric force bs zero. Explain what happens te the potential energy as two point charges of the same sign are brought together “Like charges repel. Work has to be done against the repubive forces between them to bring them closer This work is stored as electric potential energy of the system, ‘The potential energy of the two like charges therefore increases when the charges are brought closer together ELECTRIC FIELD INTENSITY/ ELECTRI € FIELD STRENGTH Electric field intensity at a point i the force experienced by a positive one coulomb charge placed in an ‘lactic field. 253Q Sner* Bat in air Enema rer? Eo = 885x107 Fm =9x10° e710 9x10°Q E=—S 5.1 unit of electric field intensity is NC™* or Vm Electric field intensity & « vector quantity and therefore direction is importemt, The direction of Eis radially outwards if the point charge is positive and radially inwards if the pint charge is negative Examples . oor wm me she ore Find Electric field intensity at P Solution em 2om 9x10°x6x107° ane Ec a a F, = 1.35x10°NC-1 towards the left Tre 135x10° —7.347x10° > yant0r poeta E = 1.2765x10°NC™? Towards left fA ODT By = 7.3472c10°NC"" towands the right 2 Ot Bsn S one me Find electric field intensity at B Solution __ 910° x3x10-6 i = 10.8x10°NC™ towards the right E = 108x108 + 4.5x10° E = 153x105NC™* Towards Right Ee _ 9x10°x5x10°8 Es =a Sx10°NC towards the right 254Find electric field intensity at B Solution Ep = 6.75x10°NC™ E= (1125x10%c0s60) + ( 6.75x10%cos60 1.125x10°sin60/ * \-6.75x10%sin60, ( 9.010" ) 38971110", (9.0x10°)? + (3.8971x10°? reo F = 981x10°NC™ 9x 10°xSx10-¢ 38971x10° oz 9.0x105 Ey = 1.125x10°NC* a=234° Fy = atau Resultant electric field is 9.81x10°NC~1at Cf 23.4° te the horizontal 4. Two point charges A and B of charges 0.10uCand 0.05,C respectively placed 8cm apart os shown below Solution m a (2ri0%coss09) Hi G Bx10%cos36, 9) 3.6x10%sin36.9) * \ 1.8x105sin36.9 (Gob) site = (145x105)? + (8.24x105)? = 355x105NCu! 9x10? x0.Ax10-6 3.24x105 4 0052 tana = 145x108 Ey = 3.6x105NC™ a= 659° Fy = zeroes Resultant electric field 6 355x10°NC™? at 005 65.89° to the horizontal Ey = 1.8x10°NC** 5, Three charges—3x10-°C , 3x10~°C and 4x10-°C are placed in a vacuum at the vertices PRO respectively at rectangle PORS of sides 3cm by 4cm as shown below 255aioe dom Sx10-%C ic field strength at $ fp = tate Ep = 1.687x10'NC~ towends left srs0Paae307* co? 10*NC~ upwards py = Selorxtet0-? a 0.05? Ey = 144x10¢NC™ 168710") 5 ( 0 ) + (Laser fcoss6.9) 0 3x10") *\1.44x10481n36.9: Sasso") E = 3.9x10*NC™ 3.865x10" 3.865x10" tana = F355 .109 i £ a= 82.11" seen Resultant electric field & 3.9210'NC~* at eseacio 82.11° tothe horizontal FY = (5355x108)? + (3.865x10")? Exercise 1. The electric field intersity at the surface of the earth is about 1.210? mat points towards the centre of the earth. Assuming that the earth is sphere of radius 6.410°m. Find the charge held by the earth surface An(5. 461 10°C)» 2. Two point charges +41 and —4uC are placed 10cm apart in ai. an ac Find the electric field intensity at point C which s a distance of 20cm form each charge. An(4,5x10° NC). The point charges A and B of charges + 0.10 1 and +0.05,1 ¢ are separated by a distance of 8.0 cm along the horizontal as shown above. Find the electric field intensity at P. An(3.55x10°NC- at 66° to horizontal). 256™ Two point charges +4.0:C and ~ 4 .oUC are separated by 10.0 cm in air as shoun below x 206m nd +4 ae loans Find the electric field intensity at point x a distance of 20.0 cm from each charge . An(4.5x10°NC-! at 75,52° to the horizontal) Two point charges +3.84C and —5.2nC are in air at points P and Q as shown belowe P=43.8nC 420m 4307 R Sem Q=-5.2nc Find the electric field intensity at R. Amn(1.89x107 NC~" at Find the electric field strength at P ° to the horizontal) 2.85 Oa tc An(142,8NC-! at 45° te the horizontal) The electric intensity at the surface of the earth is about 12 x 10" V m~' and points towards the centre of the earth. Assuming that the earth is a sphere of radius 6.4 x 10° m, find the charge held by the earth's surface, ELECTROSTATIC $HEILDING OR SCREENING It i the creation of an electrically neutral space in the neighborhood of am electric field however strong, it s by enclosing it in a hollow conductor (faraday cage) Hollow sonductor |“ The charged body is enclosed in a hollow conductor which is ecrthed. i il + Equal but opposite charge is induced on the 2 inner wails of the hollow conductor + Electric field outside will not affect the Esau charged body inside the conductor ELECTRIC FLUX 9 ‘This is the product of electric field strength at any point and area normal to the field O=AE TOTAL ELECTRIC FLUX Consider a spherical surface of radius concentric with point charge 257Q Aire? But A= 4nr? Q = 4m? 95 tar rer 0 = £| This & called Guass's law of electrostatics Guass's theorem of electrostatic states that the total electric flux passing normally through a closed surface, whatever its shape is always constant Electric field intensity due to hollow charged sphere @ Outside the sphere Sine E=— 2 Inside the sphere 'No charge resides on the inside of «hollow conductor therefore F = 0 there E x= A sraph of E against the distance of a charge trom a hollow charged sphere caver ‘ig hollow sphere ngite tne nolo Toon A sraph of E against the distance due to charges a Qo a> vom ELECTRIC POTENTIAL ‘This s the work done in moving a positive one coulomb charge from infinity to a point against an electric field. Expression for electric potential Consider +10 charge xm away from-+Q being moved from C to B through asmall displacement Ax without affecting the electric field due to +Q. +a 7 set Force on IC of charge, = 2 Work done to move the charge through Ax against the field is Aw = Total work done to bring the charge form infinity to a point a distance r from the charge of w=] Fax ro 9 [-wee 258Notes Electric potential is a scalar quantity Examples 1. hollow spherical conductor of diameter 21.4 cm carrying a charge of 6.9.x 107° Cis raised to a potential of SOY’. Find the permivity of the surrounding medium Solution Hiatt © 4ner 669x107" e= 1.026x10- Fm px10.7x10- 7x50 2. Find the potential at P onc 10cm p Solution dG y = 10x1076x9%10? V = 9x10°V Tae aaaHOAEE 3. Find the potential at B Buc 20cm Solution @ | =5x107Px9.x10° | =2.25x10°V Sear SuaanOaHE 4. Find the potential at B “uC 10cm Sem tO, 5 a = 9r10°x10-#x (24 2 V=135x10°V rer | Va oxlPxt0-%x (oat ans) 5 ia 50m =1088 sauce Find: (@) Electric potential at C () The potential energy at Cif a charge of 1ouC is placed at C (© Work done to move —10}:C charge from A toc Solution @ v=2 V, =-18x10°V ct -15_ 15 (yw =v V, = 9x10°x10-x —- =) P.e = 10x10-6x — 1.8x10° Pe 8) 259© Va = -2.7x10°V ‘9x10°x = 15x10" W=WV yp Sonea OOS Hae w= —10x10-6x(-2.7210° — Ww=9 .8x10°) Potential due te hollow charged sphere © Outside the sphere S— there V «+ GD Inside the sphere” No change resides on the inside of cs hollow conductorE = 0, therefore there no work done is done to transfer a charge from the suffice of the sphere to inside hence potential remains constant Since V = A. graph of V against the distance of a charge from a hollow charged sphere ve Tg hollow sphere insite the notion, or \Chreed aphene LT Tom A groph of V against the distance due te charges vor ee Hie aco rm ELECTRIC POTENTIAL DIFFERENCE Electric potential difference between two points is the work done to transfer +1C of charge form ‘one point to the other against cm electric field Expression for electric potential Consider two points A cnd B in an electric field which are Axm apart +0 A Force on IC of charge. = 2 Work done to move the charge through Ax against the field is Aw = Total work done to move the charge from point A to B. > w [ —Fdx a a Q -[ re * oy ay ant are 260Examples 1. Consider two points A and B at distances of 15.0 cm and 20.0 cmrespectively, from_t point charge of 6.0 jC as shoum below ----On © Find the electric potential difference between A ond B G)__Cakulate the energy required to bring a charge of + 1.0 11C from infinity to point A ° Vay = 3.6x10° ~ 2.72105 Van = 9.0210°V = 3.6x10°V @ W=0V, W = 10x10~%x3.6x10° W = 0.36) 6.0x107Fx9x10° 2 Pp | 60cm 8.4008 ~~~ Bam Sauc Find the potential energy of a charge of 25 11C placed at P Solution eo +82 Vp =Vaa+ Voo 10cm Vp = 9.6110" + 8.82x10° EHO. Vp = -7.8x10°V Gre.r P.e=QVp 98x10~%x9x10° : Pie =2.5x10~%x 78x10 ‘ng <$Gq = 8822107 Pie = -0.195) 64x10 69x10" ~ o ne 9.6x105V Vou 2613. The figure below shows point charges 44.4 uC and —22.2uC placed at the corners of a square of side 0.5m as shown below 2 ? Dom 44.4 Calculate; @_ Electric potential at 5 Gi) Potential energy of 1ouC charge placed at $ Solution o v=2 Vs = 799x105 + ~3.99x105 reer Vs = 4.0x10°7 44.Ax10-%x9xe10 eee = 7.99x10°V Gil) Pe= Qls P.e = L0x10~6x4.0x10° 2.22107 °x9.010° 05 Vs=VetVp 991 0°V P.e=4] ELECTRIC POTENTIAL GRADIENT (relation between E and V) Comider two points A and B in an electric field which are xm apart Geiy v vray A si, lf the potential at A is v and that at Bis v + Av. Then potential difference between A and B is y-Ye —(v +r) Yap = BY sro sss ones (D) Work done to move IC of change from Ato Bis equal to p.d and i given by (2) Limit os x + 0 EQUIPOTENTIAL SURFACES An equipotential surface is any two dimensional surface over which the electric potential is constant ‘and work done moving charge from one point on surface to another i zero The direction of force is always at right angles to equipotential surfaces. This implies that the és no component of electric field inside the surface Properties of equipetential surface * Work done along an equipotential surface is zero “Electric field intensity along surfaces is zero “The surfaces are at right angles to the line of force Examples 1. Calculate the electric field intensity between plates 262100v 100m ov | E 100-0 | E = 1000Vm' 2. Points A and B are 0.1m apart, « point charge of 310°°C is placed at A end emother point charge —1x10~°C is placed at B. X is « point on straight line through A and B but between A ‘and B where electric potential s zero. Calculate the distance AX Solution Aa 8 Bx10°C 7 OTK 4 xt0°C "tear 0 = 9x0", 10°%e(2 1) = 9x10?x10-%x(2- A x= 0.075m Exercise 4. Two point charge 3x10-°C and —1x10-°C are placed at points A and B respectively. A and B ore 0.2m apart amd x is « point on a straight line through A and B but between A and B. Calculate distance BX for which electric potential at xis zero, Amn(0, 1572) 2. Thefigure shows charges Q; QQ, amd Q, of —1uC,2uC,—3yC and 4uC are arranged on a straight line in vacuum a, Q Qs a (300m 20cn——»te 20cm —| (2) Cakulate potential eneray at Q: Am(—1.8x10")) (b) what is the significance of the sign of the potential energy above 3+ Alpha particles of charge 2e each having kinetic energy 1.0x10~12/ are incident head on, on a gold nuclide of charge 79e in a gold foil Caculate the distance of closest approach of an alpha paaticle and gold foil( © = 1.6x10"!C) Am(3. 64x10-"'m) 4. The figure shows charges Q; 2 and Q5 of 5uC, 64C.cmd —20uC are arranged on a straight line in Q, Q Qs a Te sem ets ooeg sane | (@) Calculate electric field intensity midway between Q; and Q; Am( 4.44x10°Vm-!) () Cakulate electric potential micway between Q, and ,Q,, An(7.85x10°V) 5. Comider two points A and C at distances of 12.0 cm and 5.0 cm respectively, from a point charge of 6.0 NC situated at B as shown below 263Calculate the energy required to bring a change of + 2.0 1:C from A to point C. An (1.261) 6. Two large oppositely charged plates are fixed 1.0 cm apart as shown below. The pd between the plates i 50V. 45%, Tem Electrof’beam } ‘An electron beam enters the region between the plates at an angle of 45° as shown, Find the maximum speed the electrons must have in order for them not to strike the upper plate [Mass of em electron = 9.11 x 10"** kg] 7. Acconducting sphere of radius 9.0 cm & maintained at an electric potential of 10kV. Cakulate tahe charge on the sphere. An(1x10"7C) Uneb 2016 (©) @ Explain an equipotential surface. (O4marks) Gi) Give an example of cm equipotential surface. (otmark) ©) (State coulomb’s lew. (©tmmerk) G3) With the aid of a sketch diagram, explain the variation of electric potential with distance from the centre of a charged metal sphere, (o3marks) Gi) Two metal plates A and B, 30cm apart are connected to a SKV d.c supply as shown below _— 7 ° Ss eset a) When a small charged sphere, 5, of mass 9.010~"kg is placed between the plates, it remains stationary. Indicate the forces acting on the sphere and determine the magnitude of the charge on the sphere. (4marks) © @ Define electric field intensity (Otmerk) Gi). With the aid of a diagram, explain electrostating shielding. (oamarks) (@) _ Explain briefly a neutral metal body is attracted to a charged body when broguth near it. (OZmearks) Uneb 2015 (@) @ Define electric potential (o1mark) Gi). Derive an expression for the electric potential at a point of a distance r, from a fixed charge. (oamarks) 264(b) With reference to a charged pecr-charged conductor. @_ Describe an experiment to show the distribution of charge on it. (03marks) Gi) Show that the surface of the conductor s cn equipotential surface. (o3martss) (© Explain how a lightning conductor protects a house from lightning. (O4marks) (@) Three charges (1, Q2 and Q; of magnitude 2:C, ~3,C, and SuC respectively are situated at comers of an equilateral triangle of sides 15cm as shown below. Qs a OG Calculate the net force on Q:. (5marks) 265CAPACITORS A capacitor isa device which stores charge A capacitor consists of a pair of oppositely charged plates separated by an insulator called a dielectric. ‘A dielectric cn insulator which brecks down when the potential difference is very high The dielectric is can be ai The symbol of a capacitor i il glass or a paper —— Charging and Discharsin: xd gp SH ee + When switch kis brought to contact x, the capacitor, ¢ charges. Current flowing through ‘the ammeter is initially high but slowly comes to zero with time when the capacitor is Fully charged + fswitch k is brought in contact with y, capacitor cs discharged. The current i initially hhigh but eventually comes to zero and in opposite direction to that when the capacitor is being charged, Explanation of charging process “When the capacitor i connected to a battery, calectrons flow from the negative terminal of the battery to the adjacent plate of the capacitor and at the scame rate electrons flow fromplate P of the capacitor towards the positive terminal of the battery leaving positive charges at P. | Positive and negative charges therefore ‘appear on the plate and oppose the flow of electrons that cause them | As charge accumulates the pad between the plates increase and charge current falls to zero ‘when the p.d between the plates of the capatitor is equal to battery voltage Explanation of discharging process th Fo Connect a wire fromthe positive plate to the negative plate. Electrons flow form the negative plate to positive plate through wire until the p.d is zero, The capacitor is fully discharged P.d against time Pd 266Note Energy changes in charging a capacitor include Chemical energy is changed to heat and electrical energy which is stored in the plates of the capacitor. Capacitance of capacitor This is the ratio of the magnitude of charge on either of the plates of a capacitor to the p.d between the plates of the capacitor Q The 5.1 unit of capacitance is farad, F Definition The farad is the capacitance of the capacitor when one coulomb of charue changes its potential difference by one vot. Examples Given the capacitance of capaxitor of 4cF and charge on the plate s 54C. Find the pd across the plate. Solution @ sxi0-* c=2 v | Y= po v 1.25 Capacitance of an isolated sphere Corsider am solated sphere of radius r. Ifthe conductor is given charge Q, then its p.cl is Q Q i Minn Gnegr 4neor =F Where &o ~ is permittivity of C= 4neor] free space Example Cakulate the capacitance of the earth given that the radius of the earth is 6.4x10m Solution C= 4negr C= 4x3,14x8, 85x10" x6.4x10° C=7.12x10-4F Capacitance of concentric spheres Consider two concentric sphere A and B each of radius a and b respectively. ogo "=a laa) po tno(52,) (Ge). ro) C= 4neo (;2) Treo Example 14 Find the capacitance of concentric spheres of radius Scm and 10cm. Given that £9 = 8.85x10- Fm“! Solution 0.1x0,09 a 0-8 C= arsareeseo® (SS) d ab c = area ( =) C= 10x10"P 2. Given two concentric sphere of radius Sem and 2cm separated by material of permittivity 8.0x10-"Fm"*, Calulate capacitance 267=a O.o2x0.05 i a ¢ = 4x3.14x8 0x10-4 (SE) 3520 NF Consider two parallel plate of capacitors each having charge Q and an area A separated by a distance d by a dielectric of permittivitye . Total electric flux @ through the surface is given by: 0 = AE. Oo) Where E is electric field intensity From Gauss law 9 = Snnennnns Equating (1) and (2) © = AE vacuum or air [c= 4] But E a Example 4. Calculate the capacitance of a parallel capacitor whose plates are 10 em by 10 cm separated b an air gap of 5 mm Solution A 5x17! 0.10.1 d SacaRTA ONS HAE C=177x10-F For parallel plate capacitor placed in 5. A parallel plate capacitor comists of two separate plates each of size 25cm and 3.0mm apart. Ifa p.d of 200V is applied to the capacitor. Calculate the charge in the plates Solution fA _@ Cc a C= Vv 185411071200 708x10-°C 5x10~12x0.25x0.25, 6+ The plates of a parallel plate capacitor each of area 2.0 cm? are 5 mm apart. The plates are in vacuum and a potential difference of 10,000V is applied across the capacitor. Find the magnitude of the charge on the capatitor. RELATIVE PERMITIVITY / DIELECRTIC CONSTANT It is defined as the ratio of capacitance of a capacitor when the insulating material (dielectric) between its plates to the capacitance of the same capacitor with a vacuum between it plates © ntl) & Gz) and ¢ == put into =e Relative permittivity can ako be defined as the ratio of the permittivity of a material to permittivity of free space Examples 268A parallel plate capacitor wos charged to 100V and then isolated. When a sheet of a dielectric is inserted between its plates, the p.d decreases to 30V. Calculate the dielectric constant of the dielectric. Solution By law of conservation of charge «, -100 Q% "30 CoVo = CV = 3.33 YC ve 2A 2uF capacitor that com just withstand a p.d of SOOOV uses a dielectric with a dielectric constant ‘6 which breaks doum if the electric field strength in it exceeds4r10” Vin. Find the: ©. Thickness of the dielectric Gi) Effective area of each plate Git) Energy stored per unit volume of dielectric Solution @ E= A=arine 5000 snegry _ 30? 4x17 = ( Sorume ~ “ae d= 125x10-%m 4 yax10-*x5000? om cat 47d Ix =e = 4246x104 6x 8852107 2A Eee bel =F 25e10F 3. A parallel plate capacitor has am arec of 00cm”, plate separation of tem and charged initially with the p.d of 100V supply, it & disconnected and a slab of dielectric 05cm thick amd relative permittivity 7 then placed between plates. @ Before the slab was inserted calculate: ) Capacitance id) Charge on the plates i Electric field strength in the gap between plates () After the dielectric was inserted, find: (0. Electric field strength Gil) P.d between the plates Gil) capacitence 8.85x107 © 7x8.85x10-2x100010-4 1.43x105Vm"! 8.85x10-PF qi) Q=cv Q = 8.85x10"12x100 100 ~ e102 Ey E = 1x10'Vm"* 7x 8,852x107 7x1 00210" at O.5x10-2 269DIELECRTIC STRENGHT It's the maximum electric field intensity an insulator ccm with stand without conducting (OF |s the maximum potential gradient on insulator can with stand without conducting USES OF DIELECTRIC It should increase the capacitance of a capacitor It s used to separate the plates of a capacitor Itreduces the chance of dielectric breakdown QUALITIES OF GOOD DIELECTRIC Itshould have a large dielectric constant + Itshould have high dielectric strength ne bo ACTION OF DIELETRIC influence but the surfaces (&*-The resuitant electric field adjacent to the plates intensity between the plates develop charge opposite to is thus reduced. But electric i fd _ the near plate, field intensity, F = “thus fH cieee qunlneans Et pad between the plates rapa reagmerssaagr al reduces, since capacitance, ie develops between the fl | __f}__s~ ecrnaisic eavanometer Hee AIM 2na14a gp-,- 2A : ) eq +) eAfV Uae ees 7Pa1= Af I 2h ee war First with ai standard capacitor of capacitance,C, suitch K; is closed and after a short time its opened Switch K, is now closed and the deflection of BG, 0, & noted ‘The capacitor bs replaced with the test capacitor of capacitance C and the procedure repeated. The deflection, 0 of BG is noted Capacitance, C i calculated from ¢ = snow connected to at The deflection 0 of the B.G is noted ‘The capacitor of capacitance C; bs then replaced by one of capacitance C2 ‘The capacitor is charged by connecting s to b. Itis then discharged through a B.G by connecting s to a. The deflection 0, of the B.G ‘s noted CO Then = & 272Q= Ut +Q But 1V,Q) = GV and Q3 = C3V Q=CV+GV GV for ¢ at +0, For capexitors connected in parallel p.d ‘across the plate of capacitors is the same (b) Capacitors in series Su | 4) Cy | ne ma y Y For capacitors connected in series charge stored ‘on the plates of capacitor the same V=V, +¥24V5 But V, =~ Examples rn +f doe] ae Lye) Nav A battery of emf 12V is connected across a system of capacitor. Calculate the total energy stored in capacitor network, Solution C= 25uF 1 Ener gy stored = CV? i Energy stored = 5x2.5x10"x12? Energy stored = 1.8x10-*] 2 ae 2u The diagram above shows a network of capacitors connected to a 120V supply. Calculate the; 273@ Charge on the au” capacitor Gi) Energy stored in 1 uP capacitor Solution uP + uP pad cross the poralle combination: V, SuP _1.0x10- © Sx10-% v= 200 Charge on au Capacitor —Qy = CV. Qa = 4107820 Q = 8.0x105C 1 8 alder c= Sr Be erate Stet aaie Total charge bar inthe circuit, Q = CV Energy stored = gttelo-'x207 Q = Fx10r4x120 Energy stored = 2x10-"J Q= 1010-46 ca tf dae] 2K ne z Lif | hoov Cakulate energy stored in a system of capacitors, i the space between the 3." éfilled with an insulator of dielectric constant 3 and capactors are fully charged Energy stored = Energy stored = 13.8x10-4) 4, A.47:°F capacitor s wed to power the flash gun of a camera. The average power output of the ‘gun is 4.0hW for a duration of the flash which is 2.Orrs. Calculate t @ Potential difference between the terminals of the capccitor immediately before a flash Gi) Maximum charge stored by the capacitor Gi) Average current provided by the capacitor during a flash Solution anaHEt 47x10-6 25835 i pear Q=274r10-2C peters? . e oe 2.74x10-* v= 583.50 eons 2x10 a= 13.74 2741 I “ht Sue : 8uF— ou ant 4 our ou our Nov The figure above shows a network of capatitors connected to « 10V batttry. Calculate the total energy stored in the network, Solution 4uFand 6uF are in series aia G 4°6 _446 1 t 1_34+12 1 Cc Pig Energy stored = zor? 4 220 0-612 1224x1047 sour ENERGY STORED IN A CAPACITOR Suppose the p.d between the plates at some Total work done to charge the capacitor to Q is instemt wos V. When a small charge of +54 is transferred from the negative plate to the positive plate, the p.d increases by dv. Work done to transfer charge, bw = (V + 6v)6q bw ~V8q ALTERNATIVELY Fromq = CV Vis proportional to q, giving a graph of v against 4 275* Work done w= average voltage x charge ro ¢ oy ee zO+NI2 “| it =30 Led rar mut Q= CV, it Areatof the shaded part =2(V +¥ + dv)5q W=5cv? work done to increase charge on the capacitor from q = 0to.q = Q JOINING Two CAPACITORS ‘When two capacitors are joined together; *% Charge flows until p.d cicross the capacitors is the some + Total charge on the circuit & conserved + Capacitors are in parallel ie C= C, + Co NOTE There is loss of eneray when capacitors are joined together. This is because charge flows until the pd across the capacitor isthe same. The flow of charge resuits in heating of the wire and hence loss in energy Examples b. A5uF capacitor is charged by a 40V supply and then connected to an un charged 20uF' capacitor. Cakulate; ©. Final p.d across each capacitor Gi) Final charge on each Gil) Energy lost Solution 5x10 F Q=25x10-%x8 Charge before= charge after connection Q=2x104€ Uat+Q=Q CW, + CV, = CV i) Energy lost = eneray before—energy after 5x10-6x40 + 20x10°%x0 = 2.521057 =(te,v,2 +1¢,v,2) Lev v=av = Gar + yCa¥2!) -70V at Qe =G 402 42 6, ) ae BxSx10™ x40? +5 x20x10°%x0' 0.0032) Q = Sx10-%x40 + 20x10-%x0 ~(fe2su05e 2x10-'C 2 © A capactor of 201.F &s connected across 50V battery supply. When it has fully charged it i then disconnected and joined to capacitor of 40 uF having a p.d of 100V. Cakulate: Effective capacitance after joining (i). The p.d on each capacitor Gil) Energy lost Solution © C=G4G i) Charge before= charge after connection C= 20x107* + 40x10-° + O2=Q C= 60x10°F CV1 4 O2V2 = 276