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

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

A
$1 .02 \times 10^{7} N / C$ upwards
B
$2 .04 \times 10^{7} N / C$ downwards
C
$2 .04 \times 10^{7} N / C$ upwards
D
$1 .02 \times 10^{7} N / C$ downwards

Solution:

$Side a = 5 \times 10^{- 2} m$

Half of the diagonal of the square $r = \frac{a}{\sqrt{2}}$

Electric field at centre due to charge q $E = \frac{kq}{{(\frac{a}{\sqrt{2}})}^{2}}$

Now field at $O = \sqrt{E^{2} + E^{2}} = E \sqrt{2} = \frac{kq}{{(\frac{a}{\sqrt{2}})}^{2}} \cdot \sqrt{2}$

$= \frac{9 \times 10^{9} \times 10^{- 6} \times \sqrt{2} \times 2}{{(5 \times 10^{- 2})}^{2}} = 1 .02 \times 10^{7} N / C (upward)$

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