Charges Q_{1} and Q_{2} are at points A and B | vedclass

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(B) Let $E_{1}$ be the electric field at $O$ due to charge $Q_{1}$ at $A$,and $E_{2}$ be the electric field at $O$ due to charge $Q_{2}$ at $B$.$E_{1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $E_{1}$ be the electric field at $O$ due to charge $Q_{1}$ at $A$,and $E_{2}$ be the electric field at $O$ due to charge $Q_{2}$ at $B$.$E_{1} = \frac{k|Q_{1}|}{x_{1}^{2}}$ along $OA$ (assuming $Q_{1}$ is positive).$E_{2} = \frac{k|Q_{2}|}{x_{2}^{2}}$ along $OB$ (assuming $Q_{2}$ is positive).Let $\angle OAB = \alpha$. Then $\angle OBA = 90^{\circ} - \alpha$.The resultant electric field $E_{net}$ is perpendicular to the hypotenuse $AB$.From the geometry of the triangle,the angle that the resultant field $E_{net}$ makes with $E_{1}$ is $\alpha$,and with $E_{2}$ is $90^{\circ} - \alpha$.Using the vector addition property,$	an(\alpha) = \frac{E_{2}}{E_{1}}$.From the triangle $OAB$,$	an(\alpha) = \frac{OB}{OA} = \frac{x_{2}}{x_{1}}$.Equating the two expressions for $	an(\alpha)$:$\frac{E_{2}}{E_{1}} = \frac{x_{2}}{x_{1}}$$\frac{k|Q_{2}| / x_{2}^{2}}{k|Q_{1}| / x_{1}^{2}} = \frac{x_{2}}{x_{1}}$$\frac{|Q_{2}| x_{1}^{2}}{|Q_{1}| x_{2}^{2}} = \frac{x_{2}}{x_{1}}$$\frac{|Q_{1}|}{|Q_{2}|} = \frac{x_{1}^{2}}{x_{2}^{2}} \cdot \frac{x_{1}}{x_{2}} = \frac{x_{1}^{3}}{x_{2}^{3}}$Thus,$Q_{1} / Q_{2}$ is proportional to $\frac{x_{1}^{3}}{x_{2}^{3}}$.

Charges $Q_{1}$ and $Q_{2}$ are at points $A$ and $B$ of a right-angled triangle $OAB$ (see figure). If the resultant electric field at point $O$ is perpendicular to the hypotenuse $AB$,then $Q_{1} / Q_{2}$ is proportional to:

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