Question

A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on the xx axis is the net electric field zero? (Be careful to keep track of the direction of the electric field of each particle.)

Answer 1

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}$

Put the value into the formula

$\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}$

$\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}$

$\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}$

$\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}$

$\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1$

$x=0.32\ m$

Hence, The position on the x axis is 0.32 m.