JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 26)
Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be -
$${{{Q_2}} \over {4\pi {\varepsilon _0}}}$$
$${{{Q^2}} \over {2\sqrt 2 \pi {\varepsilon _0}}}$$
$${{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 3 }}} \right)$$
$${{{Q_2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$$
Explanation
_10th_January_Evening_Slot_en_26_1.png)
Potential at origin = $${{KQ} \over 2} + {{KQ} \over 2} + {{KQ} \over {\sqrt {20} }} + {{KQ} \over {\sqrt {20} }}$$
(Potential at $$\infty $$ = 0)
= KQ$$\left( {1 + {1 \over {\sqrt 5 }}} \right)$$
$$ \therefore $$ Work required to put a fifth charge Q
at origin is equal to $${{{Q^2}} \over {4\pi {\varepsilon _0}}}\left( {1 + {1 \over {\sqrt 5 }}} \right)$$
Comments (0)
