Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on…

Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$ axis at which the net electric field due to these two point charges is zero is
  1. $2 \mathrm{~L}$
  2. $L / 4$
  3. $8 \mathrm{~L}$
  4. $4 \mathrm{~L}$

Solution

$-\frac{\mathrm{k} 2 q}{(\mathrm{x}-\mathrm{L})^2}+\frac{\mathrm{k} 8 \mathrm{q}}{\mathrm{x}^2}=0$ $\Rightarrow \mathrm{x}=2 \mathrm{~L}$

Asked in: JEE Main 2005

Practice more Electrostatics questions on Aicharya