Question 409: Electrostatics & Capacitance

Question

Charges Q
1
and Q
2
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 :

Accepted Answer

Net field along AB at O must be zero.
E
1
sin$$	heta $$ = E
2
cos $$	heta $$
$$ \Rightarrow $$ $${{k{Q_1}} \over {x_1^2}}.{{{x_1}} \over {AB}}$$ = $${{k{Q_2}} \over {x_2^2}}.{{{x_2}} \over {AB}}$$
$$ \Rightarrow $$ $${{{Q_1}} \over {{Q_2}}} = {{{x_1}} \over {{x_2}}}$$

Answer

$${{x_1^3} \over {x_2^3}}$$

Answer

$${{x_2^2} \over {x_1^2}}$$

Answer

$${{{x_1}} \over {{x_2}}}$$

Answer

$${{{x_2}} \over {{x_1}}}$$

Electrostatics & Capacitance

Charges Q 1 and Q 2 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 :

Choose the correct answer: