JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 10)
Charges Q1
and Q2
are at points A and B of a right angle triangle OAB (see figure). The resultant
electric field at point O is perpendicular to the hypotenuse, then
$${{{Q_1}} \over {{Q_2}}}$$ is proportional to :_6th_September_Morning_Slot_en_10_1.png)
$${{{Q_1}} \over {{Q_2}}}$$ is proportional to :
$${{x_1^3} \over {x_2^3}}$$
$${{x_2^2} \over {x_1^2}}$$
$${{{x_1}} \over {{x_2}}}$$
$${{{x_2}} \over {{x_1}}}$$
Explanation
_6th_September_Morning_Slot_en_10_2.png)
Net field along AB at O must be zero.
E1sin$$\theta $$ = E2cos $$\theta $$
$$ \Rightarrow $$ $${{k{Q_1}} \over {x_1^2}}.{{{x_1}} \over {AB}}$$ = $${{k{Q_2}} \over {x_2^2}}.{{{x_2}} \over {AB}}$$
$$ \Rightarrow $$ $${{{Q_1}} \over {{Q_2}}} = {{{x_1}} \over {{x_2}}}$$
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