Question

A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?

Answer 1

Answer:

Electric field is zero at point 4.73 m

Explanation:

Given:

Charge place = 50 cm  = 0.50 m

change q1 = 5 µC

change q2 = 4 µC

Computation:

electric field zero calculated by:

$E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2}$

Where electric field is zero,

First distance = x

Second distance = (x-0.50)

So,

E1 = E2

$k\frac{q1}{r^2}=k\frac{q2}{R^2}$

$\frac{5}{x^2}=\frac{4}{(x-50)^2}$

x = 0.263 or x = 4.73

So,

Electric field is zero at point 4.73 m