A charge q_1 = 2.32 mu c is at a distance d = 1.43 m from...

Question

Question

Answer 1

Answer #1

here,

q1 = 2.32 * 10^-6 C

q2 = -6.07 * 10^-6 C

d = 1.43 m

a)

the electric potential at (d/2) , V = k * q1 /(d/2) + k * q2/ (d/2)

V = 9 * 10^9 * 2.32 * 10^-6 /(1.43/2) - 9 * 10^9 * 6.07 * 10^-6 /(1.43/2)

V = - 4.72 * 10^-2 V

b)

let the potential is zero at x m

k * q1 /x + k * q2 /( d-x) = 0

2.32 /x - 6.07 /(1.43 -x) = 0

solving for x

x = 0.4 m

the electric potential is zero at 0.4 m

Answer 2

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