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Answer On: Question #61327, Physics / Electromagnetism

This document is a physics problem solving the charge on a small ball suspended in a horizontal electric field. It is given the mass of the ball, the electric field intensity, and the angle of suspension. The document sets up the force balance equations and solves for the charge using trigonometric relationships. It determines the charge on the ball is -3.1×10−6 C. The second part states that electric field lines cannot be discontinuous or terminate in a vacuum.

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134 views2 pages

Answer On: Question #61327, Physics / Electromagnetism

This document is a physics problem solving the charge on a small ball suspended in a horizontal electric field. It is given the mass of the ball, the electric field intensity, and the angle of suspension. The document sets up the force balance equations and solves for the charge using trigonometric relationships. It determines the charge on the ball is -3.1×10−6 C. The second part states that electric field lines cannot be discontinuous or terminate in a vacuum.

Uploaded by

jhade_cabato
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Answer on Question #61327, Physics / Electromagnetism

5 )A tiny ball of mass 0.60 g is suspended from a rigid support with a piece of thread in a
horizontal electric field of intensity 700 N/C. The ball is in equilibrium when the thread is inclined
at an angle of 20° to the vertical. What are the magnitude and sign of the charge on the ball? Take
g=9.8m/s2
a) −3.1×10−6C
b)3.2×10−6C
c)4.2×10−6C
d)−4.1×10−3
Find: q – ?
Given:
m=0.6×10-3 kg
E=700 N/C
α=20°
g=9.8m/s2
Solution:

T
f

q X

mg E

Consider the forces which acting on the tiny ball q.


Newton's Second Law:

F = ma (1)

Of (1)  T + mg + f = ma (2),

where T is tension force,


mg is gravity,
f is force of electric field
Projections of the vectors:
OX: −T sin α + f = 0 (3)
OY: T cos α − mg = 0 (4)
Force of electric field:
f = E q (5)
(5) in (3): T sin α = E q (6)
Of (4)  T cos α = mg (7)
We divide (6) on (7) term by term:
Eq
tan α = (8)
mg
mg tan α
Of (8)  q = (9)
E

Of (9)  q = 3.1 × 10−6 C


From Figure  sign of the charge: q=-3.1×10-6 C
Answer:
a) −3.1×10−6C

6) The following are true about electric field lines except that they
a) are drawn such that the magnitude of the field is proportional to the number of lines
crossing a unit area perpendicular to the lines
b) do not intersect one another
c) are discontinuous and may terminate in a vacuum
d) give the direction of motion of a unit positive test-charge under the action of the
electrostatic force
Answer:
c) are discontinuous and may terminate in a vacuum

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