 # (a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. Question:

(a) Two insulated charged copper spheres $\mathrm{A}$ and $\mathrm{B}$ have their centers separated by a distance of $50 \mathrm{~cm}$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \mathrm{C}$ ? 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?

Solution:

(a) Charge on sphere $A, q_{A}=$ Charge on sphere $B, q_{B}=6.5 \times 10^{-7} C$

Distance between the spheres, r = 50 cm = 0.5 m

Force of repulsion between the two spheres,

$F=\frac{q_{\mathrm{A}} q_{\mathrm{B}}}{4 \pi \in_{0} r^{2}}$

Where,

$\epsilon_{0}=$ Free space permittivity

$\therefore F=\frac{9 \times 10^{9} \times\left(6.5 \times 10^{-7}\right)^{2}}{(0.5)^{2}}$

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

Therefore, the force between the two spheres is $1.52 \times 10^{-2} \mathrm{~N}$.

(b) After doubling the charge, charge on sphere $A, q_{A}=$ Charge on sphere $B, q_{B}=2 \times 6.5 \times 10^{-7} C=1.3 \times 10^{-6} C$ The distance between the spheres is halved.

$\therefore r=\frac{0.5}{2}=0.25 \mathrm{~m}$

Force of repulsion between the two spheres,

$F=\frac{q_{\AA} q_{\mathrm{B}}}{4 \pi \epsilon_{0} r^{2}}$

$=\frac{9 \times 10^{9} \times 1.3 \times 10^{-6} \times 1.3 \times 10^{-6}}{(0.25)^{2}}$

$=16 \times 1.52 \times 10^{-2}$

= 0.243 N

Therefore, the force between the two spheres is 0.243 N.  