Question

If two identical conducting sphere A and B with negligible surface area have 4.5 nC and - 1.5 nC charges on them respectively. They are brought into contact once and then separated by a distance of 50 cm. The magnitude of force between them after separation will be

• A. 92.44 nN
• B. 80.89 nN
• C. 40.42 nN
• D. 242.67 nN

Answer 1

The correct option is B 80.89 nN

When spheres are brought in contact total charge gets redistributed

$Q=\frac{Q_1+Q_2}{2}=\frac{4.5-1.5}{2}=1.5nC$

After separation of 50 cm between spheres

Force | F | = $\frac{Q_1\times Q_2}{4\pi {ϵ}_0\times r^2}$

$=\frac{(1.5\times {10}^{-9})\times (1.5\times {10}^{-9})}{4\pi {ϵ}_0\times (50\times {10}^{-2}{)}^2}$

$=\frac{2.25\times {10}^{-18}}{4\pi \times 8.854\times {10}^{-12}\times (50\times {10}^{-2}{)}^2}=80.89nN$