JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 3)
Two particles A and B having charges 20$$\mu$$C and $$-$$5$$\mu$$C respectively are held fixed with a separation of 5 cm. At what position a third charged particle should be placed so that it does not experience a net electric force?
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At 5 cm from 20 $$\mu$$C on the left side of system
At 5 cm from $$-$$5 $$\mu$$C on the right side
At 1.25 cm from $$-$$5 $$\mu$$C between two charges
At midpoint between two charges
Explanation
Null point is possible only right side of $$-$$5 $$\mu$$C
_31st_August_Morning_Shift_en_3_2.png)
$${E_N} = + {{k( - 5\mu C)} \over {{x^2}}} + {{k(20\mu C)} \over {{{(5 + x)}^2}}} = 0$$
x = 5 cm
$$\therefore$$ Option (b) is correct.
$${E_N} = + {{k( - 5\mu C)} \over {{x^2}}} + {{k(20\mu C)} \over {{{(5 + x)}^2}}} = 0$$
x = 5 cm
$$\therefore$$ Option (b) is correct.
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