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Question

(C) Let $P$ be the point at distance $x$ from charge $q$ where the resultant electric field is zero.At point $P$,the electric field due to charge $q$

Vedclass · Accepted Answer

$\frac{r}{(1+\sqrt{3})}$

Vedclass · Answer

(C) Let $P$ be the point at distance $x$ from charge $q$ where the resultant electric field is zero.At point $P$,the electric field due to charge $q$ must be equal in magnitude and opposite in direction to the electric field due to charge $3q$.Thus,$\left|\vec{E}_q\right| = \left|\vec{E}_{3q}\right|$.Using the formula for electric field $E = \frac{kq}{d^2}$,we have:$\frac{kq}{x^2} = \frac{k(3q)}{(r-x)^2}$$\frac{1}{x^2} = \frac{3}{(r-x)^2}$Taking the square root on both sides:$\frac{1}{x} = \frac{\sqrt{3}}{r-x}$$r - x = \sqrt{3}x$$r = x(1 + \sqrt{3})$$x = \frac{r}{1 + \sqrt{3}}$

Two charges $q$ and $3q$ are separated by a distance $r$ in air. At a distance $x$ from charge $q$,the resultant electric field is zero. The value of $x$ is:

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