Four identical charges +50,mu C each are plac | vedclass

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(A) The side of the square is $a = 2\,m$. The distance from any corner to the centre of the square is $r = \frac{a}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \

Vedclass · Accepted Answer

$64$

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(A) The side of the square is $a = 2\,m$. The distance from any corner to the centre of the square is $r = \frac{a}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}\,m$.Potential at the centre due to one charge $q = 50\,\mu C$ is $V_1 = \frac{1}{4\pi\varepsilon_0} \frac{q}{r} = 9 	imes 10^9 	imes \frac{50 	imes 10^{-6}}{\sqrt{2}} = \frac{45 	imes 10^4}{\sqrt{2}}\,V$.Since there are four identical charges at the corners,the total potential at the centre is $V = 4 	imes V_1 = 4 	imes \frac{45 	imes 10^4}{\sqrt{2}} = 180 	imes 10^4 	imes \frac{1}{\sqrt{2}} = 90\sqrt{2} 	imes 10^4\,V$.The work done $W$ to bring a charge $q' = 50\,\mu C$ from infinity to the centre is $W = q'V$.$W = (50 	imes 10^{-6}) 	imes (90\sqrt{2} 	imes 10^4) = 4500\sqrt{2} 	imes 10^{-2} = 45\sqrt{2} \approx 45 	imes 1.414 = 63.63\,J \approx 64\,J$.

Four identical charges $+50\,\mu C$ 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 $+50\,\mu C$ from infinity to the centre of the square? (Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,\frac{N\cdot m^2}{C^2}$)

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