First slide

Potential energy

Question

Two charges $q_{1} and q_{2}$ are placed 30 cm apart, as shown in Fig. 2.135. A third charge $q_{3}$ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy I of the system is $(\frac{q_{3}}{4 {πε}_{0}}) k$ , where k is :

Easy

A

$8 q_{2}$
B

$6 q_{2}$
C

$8 q_{1}$
D

$6 q_{1}$

Solution

$U_{1} = \frac{1}{4 {πε}_{0}} (\frac{q_{1} q_{3}}{(0.4)} + \frac{q_{1} q 2}{(0.3)} + \frac{q_{2} q_{3}}{(0.5)}) U_{2} = \frac{1}{4 {πε}_{0}} (\frac{q_{1} q_{3}}{(0.4)} + \frac{q_{1} q 2}{(0.3)} + \frac{q_{2} q_{3}}{(0.1)})$
$\Rightarrow ∆ U = U_{2} - U_{1} = \frac{1}{4 {πε}_{0}} q_{2} q_{3} (\frac{1}{0.1} - \frac{1}{0.5}) = \frac{q_{3}}{4 {πε}_{0}} (8 q_{2}) \Rightarrow k = 8 q_{2}$

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