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  • Two equal positive point charges are separated by a distance 2a. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which for

Two equal positive point charges are separated by a distance 2a. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge q0 becomes maximum is   \frac{a}{\sqrt x}   The value of x is ________.

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Electric field at point “P” due to any one change   =\frac{K Q}{a^2+y^2}
 Net electric field at point “P” will be  

E_{net}=2 E \cos \alpha

\begin{aligned} & =\frac{2 \mathrm{KQ}}{\mathrm{a}^2+\mathrm{y}^2} \times \frac{\mathrm{y}}{\sqrt{\mathrm{a}^2+\mathrm{y}^2}} \\ & \mathrm{E}_{\text {net }}=\frac{2 \mathrm{KQy}}{\left(\mathrm{a}^2+\mathrm{y}^2\right)^{3 / 2}} \\ & \Rightarrow \text { Electric force }(F)=E_{\text {net }} q_0 \\ & =\frac{2 \mathrm{~K} Q q_0 \mathrm{y}}{\left(\mathrm{a}^2+\mathrm{y}^2\right)^{3 / 2}} \end{aligned}

For F = max \Rightarrow  \frac{dF}{dy}=0  

By solving, we get   y=\frac{a}{\sqrt2}

\therefore the value of x=2

 

 

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Ritika Harsh

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