Particles A and B, each carrying a charge of 2.0 nC , are at the base corners of

Question

Particles A and B, each carrying a charge of 2.0 nC , are at the base corners of an equilateral triangle 1.5 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

A) potential due to a point charge = kq/r

V = [ 9 x 10^9 x 2 x 10^-9 / 1.5 ] + [ 9 x 10^9 x 2 x 10^-9 / 1.5 ]

v = 24 volt

B) work done = q V = 5 x 10^-9 x 24 = 1.2 x 10^-7 J

C) potential energy of two charge set = kq1q2 / d

there will be 3 sets of charge .

PE = [9x 10^9 x 2 x 10^-9 x 5 x 10^-9 / 1.5 ] + [9x 10^9 x 2 x 10^-9 x 5 x 10^-9 / 1.5 ]

+ [9x 10^9 x 2 x 10^-9 x 2 x 10^-9 / 1.5 ]

PE = 1.44 x 10^-7 J

D)
V = [ 9 x 10^9 x 2 x 10^-9 / 1.5 ] + [ 9 x 10^9 x -3 x 10^-9 / 1.5 ]

= - 6 volt

E) W = -6 x 5 x 10^-9

= - 3 x 10^8 J

F) PE = [9x 10^9 x 2 x 10^-9 x 5 x 10^-9 / 1.5 ] + [9x 10^9 x 2 x 10^-9 x -3 x 10^-9 / 1.5 ]

+ [9x 10^9 x -3 x 10^-9 x 5 x 10^-9 / 1.5 ]

PE = - 6.6 x 10^-8 J

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