Question

two charges of equal magnitude and at a distance R exert a force f on each other is the charges are halved and the distance between them is doubled then the new force acting on a charge is what​

Answer 1

Hi..!!

See the above file attached.

Answer 2

If the charges are halved and the distance between them is doubled the electric force becomes (1/16)th times that of initial force.

Explanation:

The force produced between two charges is given by :

$F=k\dfrac{q_1q_2}{R^2}$

R is the distance between charges

If charges are halved,

$q_1'=\dfrac{q_1}{2}$

$q_2'=\dfrac{q_2}{2}$

Distance is doubled,

R' = 2R

New electric force is given by :

$F'=k\dfrac{q_1'q_2'}{R'^2}$

$F'=k\dfrac{(q_1/2)(q_2/2)}{(2R)^2}$

$F'=\dfrac{1}{16}\times k\dfrac{q_1q_2}{R^2}$

$F'=\dfrac{1}{16}\times F$

So, if the charges are halved and the distance between them is doubled the electric force becomes (1/16)th times that of initial force. Hence, this is the required solution.

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