Question

Two conducting spheres having charges 3Q and 5Q attract each other with a force of 10N. They are then touched with each other and then separated to the same distance again. Calculate the magnitude and nature of force now.

Answer 1

According to the Coulomb's electrostatic force(F) of attraction or repulsion between two charges is directly proportional to product of charges.

So $\mathrm{F} \propto \mathrm{q}_{1} \mathrm{q}_{2}$

The one of the charges should be negative.

a) IN the first case $q_1 = -3Q$ and $q_2 = 5Q$

Now $\mathrm{F}_1\alpha (3\mathrm{Q})(5\mathrm{Q})$

And in the second case when the two charges are touch each other then the common charge on each sphere is

$$\begin{array}{l} q = (- 3 Q + 5 Q) / 2 \\ = Q \\ \end{array}$$

Then $\mathrm{F}_2\alpha (\mathrm{Q})(\mathrm{Q})$

$$\frac {F _ {2}}{F _ {1}} = \frac {(Q) (Q)}{(3 Q) (5 Q)}$$

Then

$$= 1 / 1 5$$

Now $\mathrm{F}_2 = (1 / 15)\mathrm{F}_1$

$$= (1 / 1 5) (1 0 \mathrm {N}) = 2 / 3 \mathrm {N}$$

b)

The nature of force is repulsive