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(D) Let the electric field be zero at a point $P$ located at a distance $x$ from charge $Q$ and $(6-x)$ from charge $4Q$.At point $P$,the magnitude of

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$4$

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(D) Let the electric field be zero at a point $P$ located at a distance $x$ from charge $Q$ and $(6-x)$ from charge $4Q$.At point $P$,the magnitude of the electric field due to both charges must be equal:$\frac{KQ}{x^2} = \frac{K(4Q)}{(6-x)^2}$$\frac{1}{x^2} = \frac{4}{(6-x)^2}$Taking the square root on both sides:$\frac{1}{x} = \frac{2}{6-x}$$6 - x = 2x$$3x = 6$$x = 2 	ext{ cm}$The distance $x$ is from charge $Q$. The question asks for the distance from charge $4Q$,which is $(6-x)$.Distance from $4Q = 6 - 2 = 4 	ext{ cm}$.

Two charges $Q$ and $4Q$ are separated by a distance of $6 \text{ cm}$. The distance of the point from $4Q$ at which the net electric field is zero is: (in $\text{ cm}$)

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