A charge q1 = 2.72

A charge

q₁ = 2.72

Expert Solution

a ) electric potential at point A due to charge q1 at distance d/3( = 0.61m ) is

V1 = k*q1/r = 9*10^9*2.72*10^{-6}/0.61 = 40.13*10^3 V

electric potential at point A due to charge q2 at distance 2d/3( = 1.22m ) is

V2 = k*q2/r = 9*10^9*(-5.57)*10^{-6}/1.22 = -41.09*10^3 V

net electric potential at point A between two charges is = -960 V

b) let x be the distace from Q1 to point A

and (d-x) be the distace from Q2 to point A

then V1 = 24.42*10^{3}/x and V2 = -50.13*10^{3}/(d-x)

for net electric potential to be zero

V1 + V2 = 0

on solving we get x = 0.4521 m

hence if point is placed at a distance of 0.4521m from Q1 the electric potential is zero

genius_generous answered 4 months ago

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