Answer to Question #211250 in Electric Circuits for Cyber

Question

Two positive charges of 12×10-10C and 8×10-10C are placed 10cm apart find the work done in bringing the charges 4cm closer

Answer 1

Potential when the charges are 10cm apart

$q_1=12\times 10^{-10}C\\ q_2=8\times 10^{-10}C\\ d_1=10cm=0.1m\\ V_1=k\frac{q_1q_2}{d_1}\\ k=9\times 10^9Nm^2/C^2\\ V_1=9\times 10^9\times \frac{12\times 10^{-10}\times 8\times 10^{-10}}{0.1}\\ V_1=86.4nJ\\$

Potential when the charge is 4cm apart

$d_2=4cm=0.04m\\ V_2=k\frac{q_1q_2}{d_2}\\ k=9\times 10^9Nm^2/C^2\\ V_1=9\times 10^9\times \frac{12\times 10^{-10}\times 8\times 10^{-10}}{0.04}\\ V_2=216nJ\\$

Work done in bringing the charges 4cm closer is

$W=V_2-V_1\\ W=216-86.4=129.6nJ$