Question

Three charges $0.1$ coulomb each are placed on the corers of an equilateral triangle of side $1m$. If the energy is supplied to this system at the rate of $1kW$, how much time would be required to move one of the charges onto the midpoint of the line joining the other two?

Answer 1

Given three charges $0.1C$ placed on the corners of equilateral triangle of side $1m$. Given power $P=1kW$. We have to find the time required to move one of the charge onto the midpoint of line joining the other two.

$9A=9B=9C=0.1C$

$a=$ side of triangle $=1m$

Rate of energy supplied $=1kW$

Potential at $A$ due to charges at $B$ and $C$ is

$V_A=\frac{1}{4\pi {\epsilon }_0}[\frac{0.1}{1}+\frac{0.1}{1}]$

$=2\times 9\times {10}^9\times 0.1V$

$=18\times {10}^{8}V$

Potential at $D$ due to charges at $B$ and $C$ is

$V_D=\frac{1}{4\pi {\epsilon }_0}\frac{0.1}{0.5}+\frac{1}{4\pi {\epsilon }_0}\frac{0.1}{0.5}$

$=2\times 9\times {10}^9\times \frac{1}{5}V$

$=36\times {10}^{8}V$

Now potential difference between $A$ and $D$ is

$V_D-V_A=(36-18)\times {10}^{8}V$

$=18\times {10}^{8}V$

So ,work done in moving charge $0.1C$ from $A$ to $D$

$W=V.q=0.1C\times 18\times {10}^{8}V$

$=1.8\times {10}^{8}V$

We know that, $Power=\frac{Work}{Time}$

or, $Time=\frac{Work}{Power}=\frac{1.8\times {10}^{8}J}{1kW}$

$=\frac{1.8\times {10}^{8}J}{{10}^{3}J{s}^{-1}}$

$=1.8\times {10}^{5}sec$

So, time taken to move charge at $A$ to the mid of other two charges i.e, at $D$ is $t=1.8\times {10}^5$ seconds