Question

Two particles with positive charges q1 and q2 are separated by a distance s. Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero

Answer 1

Answer:

The distance is  $z = \frac{1 + \sqrt{ \frac{k * q_2 }{ k * q_1}}}{s}$

Explanation:

From the question we are told that

   The positive charges is  $q_1 \ and \ q_2$

   The distance of separation is  $s$

   

Generally at the point where the electric field is zero  would be a point where

The  electric field due to $q_1$ =  The  electric field due to $q_2$

Generally the electric field due to $q_1$ is mathematically represented as

     $E_1 = \frac{k * q_1}{ z^2 }$

Since we are taking our reference from $q_1$ then the electric field due to $q_2$ is mathematically represented as

     $E_2 = \frac{k * q_2}{ (s- z)^2 }$

So

     $\frac{k * q_1}{ z^2 } = \frac{k * q_2}{ (s- z)^2 }$

      $k * q_1 * (s- z)^2= k * q_2 * z^2$

      $[ \frac{s -z }{z} ]^2 = \frac{k * q_2 }{ k * q_1}$

      $\frac{s -z }{z} = \sqrt{ \frac{k * q_2 }{ k * q_1}}$

     $\frac{s}{z} - 1 = \sqrt{ \frac{k * q_2 }{ k * q_1}}$

     $z = \frac{1 + \sqrt{ \frac{k * q_2 }{ k * q_1}}}{s}$