Question: Q. 7. Two point charges $q$ and $-2 q$ are kept $d$ distance apart. Find the location of the point relative to charge $q$ at which potential due to this system of charges is zero. A [Comptt. O.D. I, II, III 2014]
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Solution:
Ans.
Let $P$ be the required point at a distance $x$ from charge $q$
$\therefore \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{x}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{(-2 q)}{(d-x)}=0$
$$ \begin{aligned} & \frac{1}{x}=\frac{2}{d-x} \ & \Rightarrow \quad x=\frac{d}{3} \end{aligned} $$
$\therefore$ The required point is at a distance $\frac{d}{3}$ from the charge $q . \quad$ [CBSE Marking Scheme 2014] $1 / 2$
