Question

Two positive charges of $20\mu C$ and $8\mu C$ are 20 cm apart. Find the work done in bringing them 5cm closer.

Answer 1

$q_1=20\times {10}^{-6}c,q_2=8\times {10}^{-6}c$

So, the primary distance $r_1=20cm=0.2m$

Final distance (20-5) = 15 = 0.15m

Now the work done = $\frac{k{q}_1{q}_2}{R}$

$=\frac{9\times {10}^9\times 20\times {10}^{-6}\times 8\times {10}^{-6}}{(0.15-0.2)}\times 0.2$

$=5.7Joule$