Three point charges Q, 4Q and 16Q are placed | vedclass

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(D) Let the charges $4Q$ and $16Q$ be at the ends,separated by a distance $d = 9 	ext{ cm}$. Let charge $Q$ be placed at a distance $x$ from $4Q$. Th

Vedclass · Accepted Answer

Both $B$ and $C$ are correct.

Vedclass · Answer

(D) Let the charges $4Q$ and $16Q$ be at the ends,separated by a distance $d = 9 	ext{ cm}$. Let charge $Q$ be placed at a distance $x$ from $4Q$. The potential energy of the system is given by:$U = \frac{k(4Q)(16Q)}{d} + \frac{k(4Q)(Q)}{x} + \frac{k(16Q)(Q)}{d-x}$To minimize $U$,we set $\frac{dU}{dx} = 0$:$\frac{dU}{dx} = -\frac{k(4Q^2)}{x^2} + \frac{k(16Q^2)}{(d-x)^2} = 0$$\frac{16}{(d-x)^2} = \frac{4}{x^2} \Rightarrow \frac{4}{d-x} = \frac{2}{x}$$4x = 2d - 2x \Rightarrow 6x = 2d \Rightarrow x = \frac{d}{3}$Given $d = 9 	ext{ cm}$,$x = \frac{9}{3} = 3 	ext{ cm}$ from $4Q$.Thus,$Q$ is $3 	ext{ cm}$ from $4Q$ and $6 	ext{ cm}$ from $16Q$. This matches option $B$.Now,check the electric field at $Q$:$E = \frac{k(4Q)}{x^2} - \frac{k(16Q)}{(d-x)^2} = \frac{k(4Q)}{3^2} - \frac{k(16Q)}{6^2} = \frac{4kQ}{9} - \frac{16kQ}{36} = \frac{4kQ}{9} - \frac{4kQ}{9} = 0$.Since both $B$ and $C$ are correct,the correct option is $D$.

Three point charges $Q, 4Q$ and $16Q$ are placed on a straight line of length $9 \text{ cm}$. The charges are placed in such a way that the system has minimum potential energy. Then:

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