First slide

Electrostatic potential energy

Question

In the rectangle shown below the two corners have charges $q_{1} = - 5 μ C$ and $q_{2} = + 2.0 μ C$ The work done in moving a charge $+ 3.0 μ C$ From $B$ to $A$ is (take $1 / 4 π ε_{0} = 10^{10} N - m^{2} / C^{2}$ )

Moderate

A

$2.8 J$
B

$2.5 J$
C

$3.2 J$
D

$3.5 J$

Solution

Work done $V_{A} = 10^{10} [\frac{(- 5 \times 10^{- 6})}{15 \times 10^{- 2}} + \frac{2 \times 10^{- 6}}{5 \times 10^{- 2}}] = \frac{1}{15} \times 10^{6} v o l t$
$a n d V_{B} = 10^{10} [\frac{(2 \times 10^{- 6})}{15 \times 10^{- 2}} - \frac{5 \times 10^{- 6}}{5 \times 10^{- 2}}] = - \frac{13}{15} \times 10^{6} v o l t ∴ W_{B A} = Q (V_{A} - V_{B}) = 3 \times 10^{- 6} [\frac{1}{15} \times 10^{6} - (- \frac{13}{15} \times 10^{6})] = 2.8 J$

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