Question

Two point charges of +3 uC and -8uC attract each other with a
force of 1 N. A charge of + 5uC is added to each of them. Now the
force will be-
(a) I N attractive,
(b) 1 N repulsive,
(c) Zero,
(d) cannot be found.
Ans.(c)​

Answer 1

Answer:The force will be 1 N attractive i.e.option(a).

Explanation:

The net force on the charges is given as,

$F=\frac{kQ_1Q_2}{d^2}$        (1)

Where,

F=force acting between the charges

k=Coulombs constant

Q₁, Q₂=charges which exert force on each other

d=distance between the charges

From the question we have,

Q₁=3μC

Q₂=-8μC

F₁=1N

Now, a charge of + 5μC is added to each of them so,

Q₁'=3μC+ 5μC=8 μC

Q₂'=-8μC+5μC =-3 μC

Now by using equation (1),

$\frac{F_1}{F_2} =\frac{Q_1Q_2}{Q_1'Q_2'}$    (2)

By substituting the required values in equation (2) we get;

$\frac{1}{F_2} =\frac{3\mu C\times -8\mu C}{8\mu C\times -3\mu C}$

$F_2=1N$

The nature of the force will be attractive since the charges are of the opposite nature.

Hence, the correct answer among the given option is 1N attractive i.e.option(a).