Particles A and B, each carrying a charge of 1.5 nC , are at the base corners of

Question

Particles A and B, each carrying a charge of 1.5 nC , are at the base corners of an equilateral triangle 3.0 m on a side.

Part A

What is the potential (relative to zero at infinity) at the apex of the triangle?

Part B

How much work is required to bring a positively charged particle carrying a charge of 5.0 nC from infinity to the apex if A and B are held fixed?

Part C

Calculate the electric potential energy for the triangular charge distribution.

Part D

What is the potential (relative to zero at infinity) at the apex of the triangle if particle B is replaced by a particle carrying a charge of 3.0 nC?

Part E

How much work is required to bring a positively charged particle carrying a charge of 5.0 nC from infinity to the apex if particle B is replaced by a particle carrying a charge of 3.0 nC and A and Bare held fixed?

Part F

Calculate the electric potential energy for the new triangular charge distribution.

Explanation / Answer

Potential due to a charge q , at distance r, from it is , kq/r
K is a constant = 9x109
Potential at a point due to number of charges is sum of potential due to each charge.
A) Potential due two charges at Apex = 2 k q /r
q = 1.5 nC and r = 3m
V = 9V  

B) W = q ( V2 - V1)
q is charge being moved from point 1 to point 2
W = 5 x 10-9 ( 9-0) = 4.5 x 10-8 J

C) Potential energy = k /r ( q1 q2 + q2q3 + q1q3)
q1, q2 and q3 are charges at three corners and r is each side of triangle
Hence potential energy U = 5.2x10-8 J

D) V = k/r (q1 + q2)
now q1 + q2 = - 1.5
Hence V = -4.5V

E) W = 5x10-9 ( -4.5 - 0) = -2.25 x 10-8 J

F) U = k/r ( -4.5 + 7.5 - 15)x10-18 = -3.6 x 10-8 J

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