Question

A charge q1 = 1.92 µC is at a distance d = 1.63 m from a second charge q2 = −5.97 µC. a) Find the electric potential at a point A between the two charges that is d/3 from q1. Note that the location A in the diagram above is not to scale. b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer 1

A and B are separated by d/3 units

if Point is a d/3 from A, then it is at 2d/3 from B

d/3 = 1.62/3 = 0.543 m

2d/3 = 0.543*2 = 1.08 m

so Potential due to each charge = Kq/r

Vnet = Va + Vb

Vnet = (9e9 * 1.92 e -6/0.543) - (9e9 * 597 e -6/1.08)

Vnet = -4.94 e 6 Volts

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since Both are oppoisde charges, net field will be zero at a point away from any of the charge

let Net EF be zero x units from 1.92 uC chage

this point is at (x - 1.63) from 5.97 uC

at this point , Edue to A = Edue to b

KQ/x^2 = KQ2/(x-1.63)^2

(x-1.63)^2/x^2 = Q2/q1 = 5.97/1.92 = 3.11

(x-1.63)/x = sqrt(3.11) = 1.763

x- 1.63 = 1.763 x

0.763 x = -1.63

x = 1.63/0.763

x = -2.13 m to the left of 1.92 uC charge, net EF is zero