Electrostatic Potential And Capacitance

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Two point charges 4Q and Q are separated by 1 m in air. At what point on the line joining the charge is the electric field intensity zero?

Solution

Given, two point charges 4Q and Q seperated by 1 m in air.
Let, the point where the electric field intensity is 0 be at a distance x from charge Q.

Therefore, $\frac{1}{4 π ε_{\circ}} \frac{Q}{x^{2}} = \frac{1}{4 π ε_{\circ}} \frac{4 Q}{(1 - x)^{2}} \frac{1}{x^{2}} = \frac{4}{(1 - x)^{2}} \frac{1}{x} = \frac{2}{1 - x} 1 - x = 2 x x = \frac{1}{3} x = 0.3 m$

Hence, at a distance 0.3 m from the first charge, electric field intensity is 0.

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Electrostatic Potential And Capacitance

Two point charges 4Q and Q are separated by 1 m in air. At what point on the line joining the charge is the electric field intensity zero?

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