Point charges q 1=+2.00 C and q 2=2.00 C are placed at adjacent corners of a squ

Question

Point charges q1=+2.00C and q2=2.00C are placed at adjacent corners of a square for which the length of each side is 4.50 cm . Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.

Part A

What is the electric potential at point a due to q1 and q2?

Part B

What is the electric potential at point b?

Part C

A point charge q3 = -4.00 C moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?

Explanation / Answer

A)
q1 = 2*10^-6 C
q2 = -2*10^-6 C
r1 = length of diagonal /2 = 4.5*sqrt(2) / 2 = 3.18 cm = 3.18*10^-2 m
r2 = 3.18*10^-2 m

Va = K* (q1/r1 + q2/r2)
= (9*10^9)*[(2*10^-6)/(3.18*10^-2) -( 2*10^-6)/(3.18*10^-2)]
= 0
Answer: 0 V

B)
q1 = 2*10^-6 C
q2 = -2*10^-6 C
r1 = length of diagonal = 4.5*sqrt(2) / 2 = 6.36 cm = 6.36*10^-2 m
r2 =side length = 4.5 cm = 4.5*10^-2 m

Vb = K* (q1/r1 + q2/r2)
= (9*10^9)*[(2*10^-6)/(6.36*10^-2) -( 2*10^-6)/(4.5*10^-2)]
= (9*10^9)* [3.14*10^-5 - 4.44*10^-5]
= -117000 V
Answer: -117000 V

C)
W = -q* (Vb-Va)
= -(4*10^-6)* (-117000 - 0)
= 0.468 J
Answer: 0.468 J

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