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 :
![](https://imagex.cdn.examgoal.net/jgy2xukfrplxdg/85fdb46f-6e0b-4e9f-99e4-987cf3d34bdd/62893f40-0462-11eb-8293-3b6cc591631c/file-jgy2xukfrplxdh-326w.png)

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 $ ${{{Q_1}} \over {{Q_2}}} = {{{x_1}} \over {{x_2}}}$
![](https://imagex.cdn.examgoal.net/jgy2xukguky9wm/1d79558a-7573-4c9b-a331-f245074092d5/bd9bb460-19c2-11eb-8293-3b6cc591631c/file-jgy2xukguky9wn-302w.png)

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: