Question

The distance between two charges $q_1-+2\mu C$ and $q_2=+8\mu C$ is $15cm.$ Calculate the distance from the charge $q_1$ to the points on the line segment joining the two charges where the electric field is zero

• A. 1 cm
• B. 2 cm
• C. 3 cm
• D. 4 cm
• E. 5 cm

Answer 1

Answer: (E)

Given, charge $q_1=+2\mu C$, charge $q_2=+8\mu C$ and distance between the charges $d=15cm$.
As, the electric field is zero at point $p$ only if the field due to charge $+2\mu C$ balances the field due to charge $+8\mu C$.
Let $x$ be the distance from charge $q_1$ at which the electric field is zero as shown in the figure below.

$\therefore \frac{k{q}_1}{x^2}=\frac{k{q}_2}{(15-x{)}^2}$
$\Rightarrow \frac{1}{4\pi {\epsilon }_0}\cdot \frac{2\times 10^{-6}}{x^2}=\frac{1}{4\pi {\epsilon }_0}\frac{8\times 10^{-6}}{(15-x{)}^2}$
$\Rightarrow \frac{1}{x^2}=\frac{4}{(15-x{)}^2}$
$\therefore \frac{1}{x}=\frac{2}{15-x}$
$\Rightarrow 2x=15-x$
$\Rightarrow x=5cm$
Thus, the required point is at a distance of $5cm$ from $2\mu C$