Question

Four point charges q, q, q and -3q are placed at the vertices of a regular tetrahedron of side L. The work done by electric force in taking all the charges to the centre of the tetrahedron is (where $k=\frac{1}{4\pi {ϵ}_0})$ :

• A. $\frac{6k{q}^2}{L}$
• B. $\frac{-6k{q}^2}{L}$
• C. $\frac{12k{q}^2}{L}$
• D. $\frac{-12k{q}^2}{L}$

Answer 1

The correct option is B $\frac{-6k{q}^2}{L}$

Initial potential energy of the system is
$U_i=k[\frac{(-3q)q}{AB}+\frac{(-3q)q}{AC}+\frac{(-3q)q}{AD}+\frac{\left(q\right)q}{BC}+\frac{\left(q\right)q}{CD}+\frac{\left(q\right)q}{BD}]$
$=k[\frac{(-3q)q}{L}+\frac{(-3q)q}{L}+\frac{(-3q)q}{L}+\frac{\left(q\right)q}{L}+\frac{\left(q\right)q}{L}+\frac{\left(q\right)q}{L}]=-\frac{6k{q}^2}{L}$
The final energy is $U_f=0$ as the potential difference at the center is zero.
Since the work done $=$ decrease in potential energy.
Work done $=-\Delta U=-(U_f-U_0)=U_0-U_f=-\frac{6k{q}^2}{L}$