Question

find the net electric field at point A in the given figure

Answer 1

Answer:

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Answer 2

Given:

q1,q2 = 2 C

d = 1m

To Find:

The electric field at point A .

Solution:

We know,

         $E=\frac{kq}{d^{2} }$

      $E1,E2 = \frac{k2}{1^{2} }$

now, electric field at point A

   $Enet = \sqrt{E1^{2}+E2^{2}+2E1E2Cos60$

            $=\sqrt{4k^{2}+4k^{2}+8k^{2}\frac{1}{2}}$

            $=\sqrt{12k^{2} }$

            $= 2k\sqrt{3\\}$  N/C

Hence, The net electric field is $2k\sqrt{3}$ N/C

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