First slide

Coulumbs law

Question

Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes

Moderate

A

$\frac{4 F}{3}$
B

$F$
C

$\frac{9 F}{16}$
D

$\frac{16 F}{9}$

Solution

In case I :
$F = - \frac{1}{4 {πε}_{0}} \frac{Q^{2}}{r^{2}} . . . . . . . . . (i) In Case II : Q_{A} = Q - \frac{Q}{4}, Q_{B} = - Q + \frac{Q}{4} ∴ F^{'} = \frac{1}{4 {πε}_{0}} \frac{(Q - \frac{Q}{4}) (- Q + \frac{Q}{4})}{r^{2}} = \frac{1}{4 {πε}_{0}} \frac{(\frac{3}{4} Q) (\frac{- 3}{4} Q)}{r^{2}} = - \frac{1}{4 {πε}_{0}} \frac{9}{16} \frac{Q^{2}}{r^{2}} . . . . . . . (ii) From equations (i) and (ii), F^{'} = \frac{9}{16} F$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App