Solveeit Logo

Question

Physics Question on Gauss Law

At the corners of an equilateral triangle of side a ( 11 meter), three point charges are placed (each of 0.1C0.1\, C ). If this system is supplied energy at the rate of 1kw1\, 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

50 h

Explanation

Solution

Initial potential energy of the system
=14πϵ0[q2a+q2a+q2a]=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{q^{2}}{a}+\frac{q^{2}}{a}+\frac{q^{2}}{a}\right]
=14πϵ0(3q3a)=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{3 q^{3}}{a}\right)
=9×109(3×(0.1)21)=9 \times 10^{9}\left(3 \times \frac{(0.1)^{2}}{1}\right)
=27×107J=27 \times 10^{7} J
Let charge at AA is moved to mid-point OO,
Then final potential energy of the system
Uf=14πϵ0(q2a2)U_{f}=\frac{1}{4 \pi \epsilon_{0}}\left(\frac{q^{2}}{a^{2}}\right)
=45×107J=45 \times 10^{7} J
Work done =UfUi=18×107J=U_{f}-U_{i}=18 \times 10^{7} J
Also, energy supplied per sec =1000J=1000\, J (given)
Time required to move one of the mid-point of the line joining the other two
t=18×1071000=18×104s=50ht=\frac{18 \times 10^{7}}{1000}=18 \times 10^{4} s =50\, h