Question

The distance between two charges $q_{1} - + 2 \mu C$ and $q_{2} = +8 \mu C$ is $15\, cm.$ 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 $d=15\,cm$.
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.

$\therefore \frac{k q_{1}}{x^{2}}=\frac{k q_{2}}{(15-x)^{2}}$
$\Rightarrow \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 \times 10^{-6}}{x^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{8 \times 10^{-6}}{(15-x)^{2}}$
$\Rightarrow \frac{1}{x^{2}}=\frac{4}{(15-x)^{2}}$
$\therefore \frac{1}{x}=\frac{2}{15-x}$
$\Rightarrow 2 x=15-x$
$\Rightarrow x=5\,cm$
Thus, the required point is at a distance of $5\,cm$ from $2\,\mu\,C$