Question

At the corners of an equilateral triangle of side a ( $1$ meter), three point charges are placed (each of $0.1\, C$ ). If this system is supplied energy at the rate of $1\, kw$, then calculate the time required to move one of the mid-point of the line joining the other two.

• A. 50 h
• B. 60 h
• C. 48 h
• D. 54 h

Answer 1

Answer: (A)

Initial potential energy of the system
$=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{q^{2}}{a}+\frac{q^{2}}{a}+\frac{q^{2}}{a}\right]$
$=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{3 q^{3}}{a}\right)$
$=9 \times 10^{9}\left(3 \times \frac{(0.1)^{2}}{1}\right)$
$=27 \times 10^{7} J$
Let charge at $A$ is moved to mid-point $O$,
Then final potential energy of the system
$U_{f}=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{q^{2}}{a^{2}}\right)$
$=45 \times 10^{7} J$
Work done $=U_{f}-U_{i}=18 \times 10^{7} J$
Also, energy supplied per sec $=1000\, J$ (given)
Time required to move one of the mid-point of the line joining the other two
$t=\frac{18 \times 10^{7}}{1000}=18 \times 10^{4} s =50\, h$