Two particles having charges - q - 1 - and - q - 2 - exert a force F on each other- when they are placed a certain distance- What will be the force between them- the distance between them is reduced to half and the charge on each particle is doubled -12 F4 F8 F16 F

Question

Toppr Admin · Accepted Answer

-F-kq1q2r2-F-x2032-kq-x2032-1q-x2032-2r-x2032-2-given- q-x2032-1-2q1-q-x2032-2-2q2-r-x2032-r2-x2234-F-x2032-F-q-x2032-1q-x2032-2q1q2r2r-x2032-2-4-xD7-4-16-F-x2032-16F-ans- -D-

Two particles having charges $q_{1}$ and $q_{2}$ exert a force $F$ on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled ?
$12 F$
$4 F$
$8 F$
$16 F$

$F = k \frac{q_{1} q_{2}}{r^{2}}$

$F^{'} = k \frac{q_{1}^{'} q_{2}^{'}}{r^{' 2}}$

given, $q_{1}^{'} = 2 q_{1}, q_{2}^{'} = 2 q_{2}, r^{'} = \frac{r}{2}$

$∴ \frac{F^{'}}{F} = \frac{q_{1}^{'} q_{2}^{'}}{q_{1} q_{2}} \frac{r^{2}}{r^{' 2}} = 4 \times 4 = 16$

$F^{'} = 16 F$

ans: (D)

Two particles having charges q1 and q2 exert a force F on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled ?12 F4 F8 F16 F

F=kq1q2r2 F′=kq′1q′2r′2 given, q′1=2q1,q′2=2q2,r′=r2 ∴F′F=q′1q′2q1q2r2r′2=4×4=16 F′=16F ans: (D)

Two particles having charges $q_{1}$ and $q_{2}$ exert a force $F$ on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled ?
$12 F$
$4 F$
$8 F$
$16 F$

$F = k \frac{q_{1} q_{2}}{r^{2}}$

$F^{'} = k \frac{q_{1}^{'} q_{2}^{'}}{r^{' 2}}$

given, $q_{1}^{'} = 2 q_{1}, q_{2}^{'} = 2 q_{2}, r^{'} = \frac{r}{2}$

$∴ \frac{F^{'}}{F} = \frac{q_{1}^{'} q_{2}^{'}}{q_{1} q_{2}} \frac{r^{2}}{r^{' 2}} = 4 \times 4 = 16$

$F^{'} = 16 F$

ans: (D)