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1. Electric Charges and Fields
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Two charges ${q_1}$ and ${q_2}$ are placed $30\,\,cm$ apart, 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
A$8\,{q_2}$
B$8\,{q_1}$
C$6\,{q_2}$
D$6\,{q_1}$
Solution

$==>$ $\Delta U = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\left( {\frac{{{q_1}{q_3}}}{{0.4}} + \frac{{{q_2}{q_3}}}{{0.1}}} \right) – \left( {\frac{{{q_1}{q_3}}}{{0.4}} + \frac{{{q_2}{q_3}}}{{0.5}}} \right)} \right]$
$==>$ $\Delta U = \frac{1}{{4\pi {\varepsilon _0}}}[8{q_2}{q_3}] = \frac{{{q_3}}}{{4\pi {\varepsilon _0}}}(8{q_2})$
$k = 8q_2$
Standard 12
Physics
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