Question

The work required to put the four charges from infinity to the position shown here is

• A. $\frac{-0.65q^2}{\pi {ϵ}_{0}a}$
• B. $\frac{-1.0q^2}{\pi {ϵ}_{0}a}$
• C. $\frac{\left(1-\frac{1}{2\sqrt{2}}\right)q^2}{\pi {\left(ϵ\right)}_{0}a}$
• D. zero

Answer 1

Answer: (A)

Work is equal to the potential energy of the system.
$U=4\frac{k\left(q\right)\left(-q\right)}{a}+2\frac{k\left(q\right)\left(q\right)}{a\sqrt{2}}$
$U=\frac{k{q}^2}{a}\left(-4+\sqrt{2}\right)$
$U=\frac{q^2}{\pi {\left(ϵ\right)}_{0}a}\left(-1+\frac{\sqrt{2}}{4}\right)=-\frac{0.65q^2}{\pi {\left(ϵ\right)}_{0}a}$