N rectangle, shown below, two corners have charges {q₁} = - 5,μ C and {q₂} = + 2.0,μ C . work

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2. Electric Potential and Capacitance

hard

$n$ the rectangle, shown below, the two corners have charges ${q_1} = - 5\,\mu C$ and ${q_2} = + 2.0\,\mu C$. The work done in moving a charge $ + 3.0\,\mu C$ from $B$ to $A$ is.........$J$ $(1/4\pi {\varepsilon _0} = {10^{10}}\,N{\rm{ - }}{m^2}/{C^2})$

$2.8$

$3.5$

$4.5$

$5.5$

Solution

(a) Work done $W = 3 \times {10^{ – 6}}({V_A} – {V_B});$ where
${V_A} = {10^{10}}\left[ {\frac{{( – 5 \times {{10}^{ – 6}})}}{{15 \times {{10}^{ – 2}}}} + \frac{{2 \times {{10}^{ – 6}}}}{{5 \times {{10}^{ – 2}}}}} \right] = \frac{1}{{15}} \times {10^6}\,volt$
and ${V_B} = {10^{10}}\left[ {\frac{{(2 \times {{10}^{ – 6}})}}{{15 \times {{10}^{ – 2}}}} – \frac{{5 \times {{10}^{ – 6}}}}{{5 \times {{10}^{ – 2}}}}} \right] = – \frac{{13}}{{15}} \times {10^6}\,volt$
$W = 3 \times {10^{ – 6}}\left[ {\frac{1}{{15}} \times {{10}^6} – \left( { – \frac{{13}}{{15}} \times {{10}^6}} \right)} \right]= 2.8 \,J$

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