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Thermal Expansion

Charges Q1 an...

Question

Charges $Q_{1}$ and $Q_{2}$ are at points $A$ and $B$ of a right angle triangle $O A B$ see figure. The resultant electric feild at point $O$ is perpendicular to the hypotenuse, then $Q_{1} / Q_{2}$ is proportional to :

A

\frac{x_{1}^{3}}{x_{2}^{3}}

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B

\frac{x_{2}}{x_{1}}

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C

\frac{x_{1}}{x_{2}}

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D

\frac{x_{2}^{2}}{x_{1}^{2}}

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Solution

The correct option is C $\frac{x_{1}}{x_{2}}$
Electric field due to charge $Q_{2} = E_{2} = \frac{k Q_{2}}{x_{2}^{2}}$
Electric field due to charge $Q_{1} = E_{1} = \frac{k Q_{1}}{x_{1}^{2}}$ From figure,
$tan θ = \frac{E_{2}}{E_{1}} = \frac{x_{1}}{x_{2}} \Rightarrow \frac{k Q_{2}}{x_{2}^{2} \times \frac{k Q_{1}}{x_{1}^{2}}} = \frac{x_{1}}{x_{2}}$
$\Rightarrow \frac{Q_{2} x_{1}^{2}}{Q_{1} x_{2}^{2}} = \frac{x_{1}}{x_{2}} \Rightarrow \frac{Q_{2}}{Q_{1}} = \frac{x_{2}}{x_{1}} o r, \frac{Q_{1}}{Q_{2}} = \frac{x_{1}}{x_{2}}$

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