Question

A -8.0 nC point charge and a +16.0 nC point charge are 12.0 cm apart on...

A -8.0 nC point charge and a +16.0 nC point charge are 12.0 cm apart on the x-axis.

A) What is the magnitude of the electric field at the point on the x-axis, between the charges, where the electric potential is zero?

Homework Answers

Answer #1

q1 = -8 nC = -8 x 10-9 C

q2 = 16 nC = 1.6 x 10-8 C

r = 12 cm = 0.12 m

Let us consider at P electric potential is 0

potential

V = kq1/x + kq2/(r-x)

V = 0

k(q1/x + q2/(r-x)) = 0

(-8 x 10-9)/x + (1.6 x 10-8)/(0.12 - x) = 0

-8/x + 1.6/(0.12 - x) = 0

-8(0.12 - x) + 16x = 0

-0.96 + 8x + 16x = 0

24x = 0.96

x = 0.04 m

Electric field at x = 0.04 m

Electric field due to q1

E1 = kq1/0.04

Electric field due to q2

E1 = kq2/0.08

net electric field

E = E1 + E2

E = k(8/0.04 + 16/0.08) x 10-9

E = 3.6 x 103 N/m

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