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.

Updated On: Jun 17, 2022
  • 50 h
  • 60 h
  • 48 h
  • 54 h
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The Correct Option is A

Solution and Explanation

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
$t=\frac{18 \times 10^{7}}{1000}=18 \times 10^{4} s =50\, h$
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Concepts Used:

Gauss Law

Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.

Gauss Law:

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.

Gauss Law Formula:

As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;

Q = ϕ ϵ0

The Gauss law formula is expressed by;

ϕ = Q/ϵ0

Where,

Q = total charge within the given surface,

ε0 = the electric constant.