Question

Three point charges q q/3 and 16q have to be arranged on positive x -axis within 20cm so that system is potential energy is minimum.Find the distance of charge q/3 from charge q.

Answer 1

Refer to the attached image ☺️

Answer 2

The distance of charge q/3 from charge q is 4 cm.

Three point charges q , q/3 and 16q have to be arranged on positive x - axis within 20 cm so that potential energy of system is minimum.

We have to find the distance of charge q/3 from charge q.

Let the distance of charge q/3 from charge q be x as shown in figure.

Potential energy of two charges separated at a distance r, is given by,

$U=\frac{Kq_1q_2}{r}$

Total potential energy = potential energy of q and q/3 + potential energy of q/3 and 16q + potential energy of 16q and q.

$U=\frac{Kq(q/3)}{x}+\frac{K(q/3)(16q)}{(20-x)}+\frac{K(16q)q}{20}$

Differentiating with respect to x, We get,

$\frac{dU}{dx}=-\frac{kq^2}{3x^2}+\frac{16kq^2}{3(20-x)^2}$

As the potential energy of the system is minimum.

$\implies\frac{dU}{dx}=0=-\frac{kq^2}{3x^2}+\frac{16kq^2}{3(20-x)^2}\\\\\implies(20-x)^2=16x^2\\\\\implies(20-x)=\pm4x\\\\\implies x=4,-20/3$

But charges are arranged on positive x-axis. so, x ≠ -20/3

hence the value of x = 4 cm.

Therefore the distance of charge q/3 from charge q is 4 cm.

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