Potential energy

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Four identical charges +50 μC each, are placed one at each corner of a square of side 2 m. How much external energy is required to bring another charge of +50 μC from infinity to the centre of the square? $(Given \frac{1}{4 π ε_{0}} = 9 \times 10^{9} \frac{{Nm}^{2}}{C^{2}})$

Solution

Potential at the centre of square
$V_{0} = 4 \times {(\frac{9 \times 10^{9} \times 50 \times 10^{- 6}}{2 / \sqrt{2}})}^{0} = 90 \sqrt{2} \times 10^{4} V$
Work done in bringing a charge (q = 50 μC) from $\infty$ to centre ( 0) of the square is $W = q (V_{0} - V_{\infty}) = {qV}_{0}$
$\Rightarrow W = 50 \times 10^{- 6} \times 90 \sqrt{2} \times 10^{4} = 64 J$

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Potential energy

Remember concepts with our Masterclasses.

Four identical charges +50 μC each, are placed one at each corner of a square of side 2 m. How much external energy is required to bring another charge of +50 μC from infinity to the centre of the square? Given 14πε0=9×109Nm2C2

Potential at the centre of squareV0=4×9×109×50×10-62/20=902×104 VWork done in bringing a charge (q = 50 μC) from ∞ to centre ( 0) of the square is W=qV0-V∞=qV0⇒W=50×10-6×902×104=64 J

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Four identical charges +50 μC each, are placed one at each corner of a square of side 2 m. How much external energy is required to bring another charge of +50 μC from infinity to the centre of the square? $(Given \frac{1}{4 π ε_{0}} = 9 \times 10^{9} \frac{{Nm}^{2}}{C^{2}})$