JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 24)
Three charges + Q, q, + Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by + Q, placed at x = 0, is zero, then value of q is :
$$-$$ $${Q \over 4}$$
+ $${Q \over 2}$$
+ $${Q \over 4}$$
$$-$$ $${Q \over 2}$$
Explanation
_9th_January_Morning_Slot_en_24_1.png)
Force on + Q charge at x = 0 due to q charge, F1 = $${{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}$$
Force on +Q charge at x = 0 due to + Q charge at x = d is,
F2 = $${{KQQ} \over {{d^2}}}$$
According to the question,
F1 + F2 = 0
$$ \Rightarrow $$ F1 = $$-$$ F2
$$ \Rightarrow $$ $${{KQq} \over {{{\left( {{d \over 2}} \right)}^2}}}$$ = $$-$$ $${{KQQ} \over {{d^2}}}$$
$$ \Rightarrow $$ 4q = $$-$$ Q
$$ \Rightarrow $$ q = $$-$$ $${Q \over 4}$$
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