Two-point charges and are separated by a distance of 1 m in the air. Calculate at which point on...

Question:

Two-point charges and are separated by a distance of 1 m in the air. Calculate at which point on the line joining the two charges is the electric potential is zero.

Coulomb's Law

From Coulomb's law which is known as an inverse-square law is the measure of force between two stationary electric charges. This is given by the form

$$\begin{align*} F=k\frac{q_1q_2}{r^2} \end{align*} $$ where {eq}\begin{align*} k,q_1,q_2,r \end{align*} {/eq} is Coulomb's constant, the two point charges, and the separation distance respectively. The electric field in this manner follows to be

$$\begin{align*} E=k\frac{q}{r^2} \end{align*} $$

Answer and Explanation:

Two point charges are separated by a distance of 1 meter in air. We let the point charges be {eq}\begin{align*} q_1=q \end{align*} {/eq} and {eq}\begin{align*} q_2=q \end{align*} {/eq}. There is a point along the line joining the two charges where the potential is zero. With this, we let {eq}\begin{align*} x \end{align*} {/eq} be the distance from {eq}\begin{align*} q_1 \end{align*} {/eq} and {eq}\begin{align*} 1-x \end{align*} {/eq} be the distance from {eq}\begin{align*} q_2 \end{align*} {/eq}. Using Coulumb's law, the electric field due to these two point charges are equated (following Newton's third law) to get the distance

$$\begin{align*} \frac{kq_1}{x^2}&=\frac{kq_2}{(1-x)^2}&\text{[ We assume that the charges are identical so }q_1=q_2=q \text{]}\\\\ \frac{kq}{x^2}&=\frac{kq}{(1-x)^2}&\text{[Cancel out q,k and rearrange]}\\\\ \frac{1}{x^2}&=\frac{1}{(1-x)^2}&\text{[Take the square-root and solve for x]}\\\\ x&=1-x&\text{[Solve for x]}\\\\ x&=\boxed{\frac{1}{2}~m} \end{align*} $$