Question

- two charges of equal magnitudes and at a distance r exert a force on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is

Answer 1

Answer:

F/16

Explanation:

Answer 2

Answer:

Two charges of equal magnitudes and at a distance r exert a force on each other. If the charges are halved and the distance between them is doubled, then the new force acting on each charge is $\frac{F_{1} }{16}$

Explanation:

When two charges  $q_{1}$ and $q_{2}$ are separated by a distance $r$ they will interact with each other, the magnitude of the force is given by the expression

$F_{1}=\frac{Kq_{1} q_{2} }{r^{2} }$

since both charges are equal $F_{1}=\frac{Kq^{2} }{r^{2} }$

from the question, both charges are halved and the distance between them is doubled, that is $q_{1} =q_{1} /2$, $q_{2} =q_{2}/2$,$r=2r$

now the new force is obtained by substituting these values in the force equation so, the new force

 $F_{2}=\frac{Kq_{1}/2* q_{2}/2 }{4r^{2} }$

       $=\frac{Kq_{1} q_{2} }{16r^{2} }$

       $=\frac{Kq^{2} }{16r^{2} }$

$F_{2} =F_{1}/16$

the new force is $1/16th$ the first force that much reduced