Question

Two tiny metal spheres A and B of mass mA 5 g and mB 10 g have equal positive charges q = 5 UC. The spheres are connected by a massless nonconducting string of length d = 1 m, which is much greater than the radii of the spheres as shown in fig (a). Now, Suppose you cut the string as shown in fig (b). At that instant, what is the acceleration of sphere A (in m/s")? WA We 1.00 m www () W Answer:

Answer 1

Solution:

Given data

The mass of sphere A, $m_{A}=5\; \mathrm{gm}=5\times 10^{-3}\; \mathrm{kg}$

The mass of sphere B, $m_{B}=10\; \mathrm{gm}=10\times 10^{-3}\; \mathrm{kg}$

These two spheres are separated by a distance, $d=1\; \mathrm{m}$

Charge on sphere A, $q_{A}=5\; \mathrm{\mu C}=5\times 10^{-6}\; \mathrm{C}$

Charge on sphere B, $q_{B}=5\; \mathrm{\mu C}=5\times 10^{-6}\; \mathrm{C}$

As the two spheres are having a similar charge (+ve), once the string cut, there will be a repulsive force between them.

First, we calculate this repulsive force as follows

$F=\frac{1}{4\pi \varepsilon _{o}}\frac{q_{A}q_{B}}{d^{2}}$

Now, substitute all the known values in this formula

$F=\frac{1}{(4\times 3.14\times 8.85\times 10^{-12})}\frac{(5\times 10^{-6}\times 5\times 10^{-6})}{1^{2}}$

$F=(9\times 10^{9})\times \frac{(25\times 10^{-12})}{1^{2}}$

$F=225\times 10^{-3}$

$F=0.225\; \mathrm{N}$

This is the repulsive force acting between the metal spheres A and B. When we cut the string, they start move away from each other due to this repulsive force.

Let us consider the sphere A gained an acceleration $a_{A}$

As per Newton's second law we have

where

$m_{A}=\mathrm{mass\; of\; the\; sphere\; A}=5\times 10^{-3}\; \mathrm{kg}$

$a_{A}=\mathrm{Acceleration\; of\; the\; sphere\; A}$

Now, substitute the known values in the below equation

$a_{A}=\frac{0.225}{m_{A}}$

$a_{A}=\frac{0.225}{5\times 10^{-3}}$

${\color{Blue} a_{A}=45\; \mathrm{m/s^{2}}}$