Question

Two electrically charged particles having charges of different magnitude when placed at a distance d from each other experience a force of attraction F. These two particles are put in contact and again placed at the same distance from each other. What is the nature of new force between them? Is the magnitude of force of interaction between them now more or less than F?

Answer 1

Answer:

Force between two charges is of repulsion type and the magnitude is more than the initial magnitude now

Explanation:

When two charge particles are at distance d from each other they both experience attraction force

So the two charges are of opposite sign

and the force between them is given as

$F = \frac{kq_1q_2}{d^2}$

now when they are in contact with each other then the total charge is divided equally on them

so we have

$q = \frac{q_1 - q_2}{2}$

so we have

$F' = \frac{k(\frac{q_1 - q_2}{2})^2}{d^2}$

now we have

$\frac{F}{F'} = \frac{4q_1q_2}{(q_1 - q_2)^2}$

here we know that

$(q_1 - q_2)^2 > 4q_1q_2$

so we have

$\frac{F}{F'} < 1$

$F' > F$

So force will be more than initial force and it is of repulsion type force now

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Topic : Electrostatics

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