Question

Charges $4Q$ and $Q$ are kept separated by a certain distance and are placed at $A$ and $B$, respectively. Electric intensity at $B$ is $E$. Electric intensity at $A$ is

• A. $4E$
• B. $\frac{E}{4}$
• C. $\frac{-E}{4}$
• D. $- 4 E$

Answer 1

Answer: (C)

$\frac{-E}{4}$

Answer 2

Electric intensity at $B = E$
$= \frac{4Q}{4 \pi \varepsilon_0 (AB)^2 }$
Electric field intensity at A,
$E' = \frac{Q}{4 \pi \varepsilon_0 (AB)^2 } = \frac{E}{4}$
Direction of $\vec{E}$ is opposite to that of $\vec{E}$.
So, $\vec{E}' = - \frac{\vec{E}}{4}$