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 to D. The change in the potential energy of the system is $\frac{q_{3}}{4\pi\varepsilon_{0}}\ k,$ where k is [image:2_image83.emf]

Question

Dr. Abubacker · Accepted Answer

(a) When charge $q_{3}$is at $C$, then its potential energy is
$U_{C} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{q_{1}q_{3}}{0.4} + \frac{q_{2}q_{3}}{0.5} ight)$
Where charge $q_{3}$is at $D$, then
$U_{D} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{q_{1}q_{3}}{0.4} + \frac{q_{2}q_{3}}{0.1} ight)$
Hence, change in potential energy
$\mathrm{\Delta}U = U_{D} - U_{C}$
$= \frac{1}{4\pi\varepsilon_{0}}\left( \frac{q_{2}q_{3}}{0.1} - \frac{q_{2}q_{3}}{0.5} ight)$
$but\ \ \ \ \mathrm{\Delta}U = \frac{q_{3}}{4\pi\varepsilon_{0}}k\ $
$	herefore\frac{q_{3}}{4\pi\varepsilon_{0}}k = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{q_{2}q_{3}}{0.1} - \frac{q_{2}q_{3}}{0.5} ight)$
$\Longrightarrow k = q_{2}(10 - 2) = 8q_{2}$

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 to D. The change in the potential energy of the system is $\frac{q_{3}}{4\pi\varepsilon_{0}}\ k,$ where k is

Explanation

Reviewed by Dr. Abubacker

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