Four charges are placed on corners of a squar | vedclass

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Question

(a) Side $a = 5 	imes 10^{-2} m$
Half of the diagonal of the square $r = \frac{a}{{\sqrt 2 }}$
Electric field at centre due to charge q $E = \frac{

Vedclass · Accepted Answer

$1.02 	imes {10^7}N/C$ upwards

Vedclass · Answer

(a) Side $a = 5 	imes 10^{-2} m$
Half of the diagonal of the square $r = \frac{a}{{\sqrt 2 }}$
Electric field at centre due to charge q $E = \frac{{kq}}{{{{\left( {\frac{a}{{\sqrt 2 }}} ight)}^2}}}$
Now field at O $ = \sqrt {{E^2} + {E^2}} = E\sqrt 2 $$ = \frac{{kq}}{{{{\left( {\frac{a}{{\sqrt 2 }}} ight)}^2}}}.\sqrt 2 $
$ = \frac{{9 	imes {{10}^9} 	imes {{10}^{ - 6}} 	imes \sqrt 2 	imes 2}}{{{{(5 	imes {{10}^{ - 2}})}^2}}}$$ = 1.02 	imes {10^7}\,N/C$ (upward)

Four charges are placed on corners of a square as shown in figure having side of $5\,cm$. If $Q$ is one microcoulomb, then electric field intensity at centre will be

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