JEE MAIN - Physics (2020 - 7th January Morning Slot - No. 14)
Two infinite planes each with uniform surface charge density to are kept in such a way that the angle between them is 30o. The electric field in the region shown between them is given by :
_7th_January_Morning_Slot_en_14_1.png)
$${\sigma \over {{ \in _0}}}\left[ {\left( {1 + {{\sqrt 3 } \over 2}} \right)\widehat y + {{\widehat x} \over 2}} \right]$$
$${\sigma \over {2{ \in _0}}}\left[ {\left( {1 + \sqrt 3 } \right)\widehat y + {{\widehat x} \over 2}} \right]$$
$${\sigma \over {2{ \in _0}}}\left[ {\left( {1 + \sqrt 3 } \right)\widehat y - {{\widehat x} \over 2}} \right]$$
$${\sigma \over {2{ \in _0}}}\left[ {\left( {1 - {{\sqrt 3 } \over 2}} \right)\widehat y - {{\widehat x} \over 2}} \right]$$
Explanation
_7th_January_Morning_Slot_en_14_2.png)
Electric field due to each sheet is uniform and equal to E = $${\sigma \over {2{\varepsilon _0}}}$$
$$ \therefore $$ Electric field along y direction
= $$\left( {{\sigma \over {2{\varepsilon _0}}} - {\sigma \over {2{\varepsilon _0}}}\sin 60^\circ } \right)\widehat y$$
Also Electric field along x direction
= - $${{\sigma \over {2{\varepsilon _0}}}\cos 60^\circ \widehat x}$$
$$ \therefore $$ Now net electric field between plates
= - $${{\sigma \over {2{\varepsilon _0}}}\cos 60^\circ \widehat x}$$ + $$\left( {{\sigma \over {2{\varepsilon _0}}} - {\sigma \over {2{\varepsilon _0}}}\sin 60^\circ } \right)\widehat y$$
= $${ - {\sigma \over {4{\varepsilon _0}}}\widehat x}$$ + $${\sigma \over {2{\varepsilon _0}}}\left( {1 - {{\sqrt 3 } \over 2}} \right)\widehat y$$
= $${\sigma \over {2{ \in _0}}}\left[ {\left( {1 - {{\sqrt 3 } \over 2}} \right)\widehat y - {{\widehat x} \over 2}} \right]$$
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