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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 :

Answer:

The distance is  [tex]z = \frac{1 + \sqrt{ \frac{k * q_2 }{ k * q_1}}}{s}[/tex]

Explanation:

From the question we are told that

   The positive charges is  [tex]q_1 \ and \ q_2[/tex]

   The distance of separation is  [tex]s[/tex]

   

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

The  electric field due to [tex]q_1[/tex] =  The  electric field due to [tex]q_2[/tex]

Generally the electric field due to [tex]q_1[/tex] is mathematically represented as

     [tex]E_1 = \frac{k * q_1}{ z^2 }[/tex]

Since we are taking our reference from [tex]q_1[/tex] then the electric field due to [tex]q_2[/tex] is mathematically represented as

     [tex]E_2 = \frac{k * q_2}{ (s- z)^2 }[/tex]

So

     [tex]\frac{k * q_1}{ z^2 } = \frac{k * q_2}{ (s- z)^2 }[/tex]

      [tex]k * q_1 * (s- z)^2= k * q_2 * z^2[/tex]

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

      [tex]\frac{s -z }{z} = \sqrt{ \frac{k * q_2 }{ k * q_1}}[/tex]

     [tex]\frac{s}{z} - 1 = \sqrt{ \frac{k * q_2 }{ k * q_1}}[/tex]

     [tex]z = \frac{1 + \sqrt{ \frac{k * q_2 }{ k * q_1}}}{s}[/tex]

     

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