Question

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

• A. $\frac{- 0 .65 q^{2}}{\pi \epsilon _{0} a}$
• B. $\frac{- 1 .0 q^{2}}{\pi \epsilon _{0} a}$
• C. $\frac{\left(1 - \frac{1}{2 \sqrt{2}}\right) q^{2}}{\pi \left(\epsilon \right)_{0} a}$
• D. zero

Answer 1

Answer: (A)

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