We are given a square of a side 1m with two charges at the corners shown. Q1=-(4

Question

We are given a square of a side 1m with two charges at the corners shown.
Q1=-(4/9) ×10^-9 c
Q1=+(4/9) ×10^-9 c

a) What is the electric potential at point A?

My teachers answer was -3v

b)If we place a charge of -2c at point A, what is the potential energy of this charge?

My teachers answer was 6J

c) How much work is done by the two original charges on this third charge -2c when this third charge moves from point A to infinity?

My teachers answer was 6J

Please explain/show me how he arrived at these answers? I will vote up answer and thank you!

Explanation / Answer

Given,
q1 = - 4/9 * 10^-9 C
q2 = + 4/9 * 10^-9 C
a = 1 m
(a)
Electric Potential at Point A, Va = k*q1/a + k*q2/a
As the two charges are opposite and equal , Va = 0

(b)
As Potential at Point A, is Zero
Potential Energy, U = q*Va = 0

(c)
At Initial position Potential Energy = 0,
At infinity Potential Energy = 0,
So No work is done

Either your Teacher answers are all wrong or you have copied wrong numbers !!!

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