Question

Two positive charges $2\mu C$ and 10$\mu C$ are initially Separated by 10cm. The work done in bringing the charges 4cm closer

• A. 7.2J
• B. 3.6J
• C. 8.4J
• D. 12.4J

Answer 1

Answer: (A)

Work = change in P.E
$=k\cdot \frac{9_1{9}_2}{r}$
change in $PE=k\cdot q_1{q}_2\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$
$=k\cdot q_1{q}_2\left[\frac{1}{0.06}-\frac{1}{0.1}\right]$
$=9\times 10^9\times 12\times 10^{-6}\times 10\times 10^{-6}[16.67-10]$
$=9\times 10^9\times 12\times 10^{-6}\times 10\times 10^{-6}\times 6.67$
$=7.2J$