Two identical charges $q_{1}$ and $q_{2}$ each at $1 μ C$ are fixed $30 c m$ apart as shown in the figure. A third identical charge $q_{3}$ is moved along the arc of a circle of radius $40 c m$ from $C$ to $D$ is work done in this process by external agent:

144 J

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72 m J

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288 m J

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- 144 m J

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Solution

The correct option is A $144 J$
Two identical charges $q_{1}$ and $q_{2}$ $1 μ e$ (field)
$d = 30 c m = 30 \times 10^{- 2} m$
& third identical charges $= q_{3}$
radius $= 40 c m$ from $C$ to $D$ .
$W =$
The charges in potential energy of the system is
$V_{D} - V_{C}$ is discussed under when charge $q_{3}$ is at $C$ ,
then its potential energy is $U_{C} = \frac{1}{4 π ϵ_{0}} (\frac{q_{1} q_{2}}{0.4} + \frac{q_{2} q_{3}}{0.5})$
When charges $q_{3}$ is at $D$ , then
$U_{D} = \frac{1}{4 π ϵ_{0}} (\frac{q_{1} q_{3}}{0.4} + \frac{q_{2} q_{3}}{0.1})$
hence change in potential energy
$Δ_{U} = U_{D} - U_{C}$
$= \frac{1}{4 π ϵ_{0}} (\frac{q_{2} q_{3}}{0.1} - \frac{q_{2} q_{3}}{0.5})$
but
$Δ U = \frac{q_{3}}{4 π ϵ_{0}} K$
$∴$ $\frac{q_{3}}{4 π ϵ_{0}} K = \frac{1}{4 π ϵ_{0}} (\frac{q_{2} q_{3}}{0.1} - \frac{q_{2} q_{3}}{0.5})$
$K = q_{2} (10 - 2)$
$= 8 q_{2} = 8 \times 18 = 144 J$

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Similar questions

Q. Two charges

q_{1}

and

q_{2}

are placed

30 cm

apart, as shown in the figure. A third charge

q_{3}

is moved along the arc of a circle of radius

40 cm

from

C

D

. The change in the potential energy of the system is

\frac{q_{3}}{4 π ϵ_{0}} k

, where

k

Two charges $q_{1}$ and $q_{2}$ are placed 30cm apart, as shown in figure. A third charges $q_{3}$ is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is $\frac{q_{3}}{4 π ϵ_{0}} k$ , where k is