Question

(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times {10}^{-7}$ C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer 1

$\text{Part (a):}$

$\text{Step 1: Electrostatic repulsion force Initially}$

$Q_1=Q_2=6.5\times {10}^{-7}C,r=50cm=0.5m$

By Coulomb's Law:

$F=\frac{K{Q}_1{Q}_2}{r^2}=\frac{9\times {10}^9\times 6.5\times {10}^{-7}C\times 6.5\times {10}^{-7}C}{(0.5m{)}^2}$

$F=1.52\times {10}^{-2}N$

$\text{Part (b):}$

$\text{Step 2: Electrostatic force in second case}$

When charges are doubled and distance is halved.

So, $Q_1^{\prime }=Q_2^{\prime }=2\times 6.5\times {10}^{-7}C,\&r^{\prime }=25cm=0.25m$

$F^{\prime }=\frac{K{Q}_1^{\prime }{Q}_2^{\prime }}{(r^{\prime }{)}^2}=\frac{9\times {10}^9\times (2\times 6.5\times {10}^{-7}C{)}^2}{(0.25m{)}^2}$
$F^{\prime }=0.24N$