Two charges $Q$ and $4 Q$ are separated by a distance of $6 \mathrm{~cm}$. The distance of the point from $4…

Two charges $Q$ and $4 Q$ are separated by a distance of $6 \mathrm{~cm}$. The distance of the point from $4 Q$ at which net electric field is zero is
  1. $2 \mathrm{~cm}$
  2. $6 \mathrm{~cm}$
  3. $8 \mathrm{~cm}$
  4. $4 \mathrm{~cm}$

Solution


Let the electric field be zero at a point $P$. $\begin{aligned} & \frac{K Q}{x^2}=\frac{K 4 Q}{(6-x)^2} \quad\left(\because E=\frac{K Q}{r^2}\right) \\ & (6-x)^2=4 x^2 \Rightarrow 6-x=2 x \\ & \Rightarrow x=2 \mathrm{~cm}\end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

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