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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
$1.02 \times {10^7}N/C$ upwards
$2.04 \times {10^7}N/C$ downwards
$2.04 \times {10^7}N/C$ upwards
$1.02 \times {10^7}N/C$ downwards
Solution
(a) Side $a = 5 \times 10^{-2} m$
Half of the diagonal of the square $r = \frac{a}{{\sqrt 2 }}$
Electric field at centre due to charge q $E = \frac{{kq}}{{{{\left( {\frac{a}{{\sqrt 2 }}} \right)}^2}}}$
Now field at O $ = \sqrt {{E^2} + {E^2}} = E\sqrt 2 $$ = \frac{{kq}}{{{{\left( {\frac{a}{{\sqrt 2 }}} \right)}^2}}}.\sqrt 2 $
$ = \frac{{9 \times {{10}^9} \times {{10}^{ – 6}} \times \sqrt 2 \times 2}}{{{{(5 \times {{10}^{ – 2}})}^2}}}$$ = 1.02 \times {10^7}\,N/C$ (upward)
