JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 13)
What will be the magnitude of electric field at point O as shown in the figure? Each side of the figure is l and perpendicular to each other?
_27th_July_Evening_Shift_en_13_1.png)
$${1 \over {4\pi {\varepsilon _0}}}{q \over {{l^2}}}$$
$${1 \over {4\pi {\varepsilon _0}}}{q \over {(2{l^2})}}\left( {2\sqrt 2 - 1} \right)$$
$${q \over {4\pi {\varepsilon _0}{{(2l)}^2}}}$$
$${1 \over {4\pi {\varepsilon _0}}}{{2q} \over {2{l^2}}}\left( {\sqrt 2 } \right)$$
Explanation
$${E_1} = {{kq} \over {{l^2}}} = {E_2}$$
$${E_3} = {{kq} \over {{{(\sqrt 2 l)}^2}}} = {{kq} \over {2{l^2}}}$$
$$E = {{\sqrt 2 kq} \over {{l^2}}} - {{kq} \over {2{l^2}}} = {{kq} \over {2{l^2}}}\left( {2\sqrt 2 - 1} \right)$$
_27th_July_Evening_Shift_en_13_2.png)
$${E_3} = {{kq} \over {{{(\sqrt 2 l)}^2}}} = {{kq} \over {2{l^2}}}$$
$$E = {{\sqrt 2 kq} \over {{l^2}}} - {{kq} \over {2{l^2}}} = {{kq} \over {2{l^2}}}\left( {2\sqrt 2 - 1} \right)$$
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