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The document contains a series of physics problems and solutions related to electrostatics and charge distribution. It includes calculations for the force between two charges, charge enclosed by a cuboidal box, total charge within a specified volume, linear charge density of a rod, and a brief historical note on Charles-Augustin de Coulomb, the inventor of Coulomb's law. Each problem is followed by its respective solution with detailed calculations.

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Marvin Yap
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34 views5 pages

Quiz Emags

The document contains a series of physics problems and solutions related to electrostatics and charge distribution. It includes calculations for the force between two charges, charge enclosed by a cuboidal box, total charge within a specified volume, linear charge density of a rod, and a brief historical note on Charles-Augustin de Coulomb, the inventor of Coulomb's law. Each problem is followed by its respective solution with detailed calculations.

Uploaded by

Marvin Yap
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Questions:

 A charge Q1 = 810–6 C at M (2,5,8) and a charge of Q2 = –510–6


C at N (6,15,8) are located in a vacuum. Determine the force
exerted on Q2 by Q1.

 A cuboidal box penetrates a huge plane sheet of charge with


uniform surface charge density 2.5 ×1 0−2such that its smallest
surfaces are parallel to the sheet of charge. If the dimensions of
the box are 10 cm × 5 cm × 3 cm, then find the charge enclosed by
the box.

 Find the total charge inside the volume indicated by ρ v = 4xyz2, 0 ≤


ρ ≤ 2, 0 ≤ Φ ≤ π/2, 0 ≤ z ≤ 3. All values are in SI units.

 A long thin rod of length 50 cm has a total charge of mC uniformly


disturbed over it. Find the linear charge density.

 He’s a French army Engineer who performed an elaborate series


of experiments using devices. He is the one who invented the
Coulomb’s law.
Answers:

1. R= r2-r1
= (6ax, 15ay, 8az) – (2ax, 5ay, 8az)
= (4ax + 10ay)
R
a= ¿ R∨¿ ¿
= √ ¿ ¿ = 10.77

( 4 a x + 10 a y )
= 10.77

1 Q1 Q2
F= 4 π ε0 ¿ ¿

−6 −6
1 (8 ×1 0 )(−5 ×1 0 )
¿ −12
4 π (8.854 ×1 0 ) ¿¿

( 4 a x +10 a y )
= (−3.1 ×1 0−3 )( 10.77
)

= (−1.15 a x −2.88 a y ) mN
2.

= 5 cm × 3 cm = 15 cm2
= 15 × 10–4 m2

Charge density;
σ= 2.5 ×10–2 Cm–2
Charge enclosed;
q=σA
= (2.5×10−2) (15×10−4)

= 37.5×10−6C
= 37.5µC

3.
2
ρ v =4 ⋅ ρsin φ⋅ ρ cos φ ⋅ z

π
3 2 2
Q=∫ ρv dv
vol
¿∫ ∫ ∫ (4 ⋅ ρ sin φ ⋅ ρ cos φ ⋅ z 2 )(dρ ⋅ ρdφ⋅ dz)
4.
q=5
mC = 5 × 10–3 C
l= 50 cm = 0.5 m.
λ =?

q
λ= l

−3
(5 ×1 0 )
= 0.5 = 10−2 Cm−1

5.

Charles – Augustin de Coulomb

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