As shown in the figure,at what distance in cm | vedclass

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(C) Let the point where the electric field is zero be at a distance $x$ from the charge $10 \, mC$ (point $A$).The electric field due to charge $q_1 =

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$33$

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(C) Let the point where the electric field is zero be at a distance $x$ from the charge $10 \, mC$ (point $A$).The electric field due to charge $q_1 = 10 \, mC$ at distance $x$ is $E_1 = \frac{k q_1}{x^2}$.The electric field due to charge $q_2 = 20 \, mC$ at distance $(80 - x)$ is $E_2 = \frac{k q_2}{(80 - x)^2}$.For the electric field to be zero,$E_1 = E_2$,so $\frac{q_1}{x^2} = \frac{q_2}{(80 - x)^2}$.Taking the square root on both sides: $\frac{\sqrt{q_1}}{x} = \frac{\sqrt{q_2}}{80 - x}$.Substituting the values: $\frac{\sqrt{10}}{x} = \frac{\sqrt{20}}{80 - x}$.$\frac{1}{x} = \frac{\sqrt{2}}{80 - x} \implies 80 - x = x\sqrt{2}$.$80 = x(1 + \sqrt{2}) \implies x = \frac{80}{1 + 1.414} = \frac{80}{2.414} \approx 33.14 \, cm$.

As shown in the figure,at what distance in $cm$ from point $A$ will the electric field be zero?

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