Two charges q_1 and q_2 are placed 30,cm apar | vedclass

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Question

$ \mathrm{U}_{\mathrm{i}}= \frac{1}{4 \pi \epsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{(0.4)}+\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{(0.3)}+\

Vedclass · Accepted Answer

$8q_2$

Vedclass · Answer

$ \mathrm{U}_{\mathrm{i}}= \frac{1}{4 \pi \epsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{(0.4)}+\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{(0.3)}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{(0.5)}\right] $

$ \mathrm{U}_{\mathrm{f}}=\frac{1}{4 \pi \epsilon_{0}}\left[\frac{\mathrm{q}_{1} \mathrm{q}_{3}}{(0.4)}+\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{(0.3)}+\frac{\mathrm{q}_{2} \mathrm{q}_{3}}{(0.1)}\right] $

There fore $ \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}=\frac{1}{4 \pi \mathrm{\epsilon}_{0}} \mathrm{q}_{2} \mathrm{q}_{3}\left(\frac{1}{0.1}-\frac{1}{0.5}\right) $

$=\frac{\mathrm{q}_{3}}{4 \pi \epsilon_{0}}\left(8 \mathrm{q}_{2}\right) $

$ \Rightarrow \mathrm{K}=8 \mathrm{q}_{2} $

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 $\frac{{{q_3}}}{{4\pi \,{ \in _0}}}k$ , where $k$ is

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