What can you say about two charges…?

q1 and q2, if the electric field one fourth of the way from q1 to q2 is zero?

The answer is that the ratio of q1 and q2 is 1:9. How did they get this? Did the answer relate, somehow to this equation- E= k.q1.q2/r^2 ?


2 Answers

  • Absolutely.

    Consider 'r' as the distance between the two charges q1 and q2.

    The force experienced by a unit charge (+ coulomb) at point P due to q1 is k.q1/d^2 .

    Let the point P be at a distance of 'r'/4 from q1 along the line joining q1 and q2. Then it will be at a distance of 3r/4 from q2.

    Hence 'F1' = k.16.q1.(1)/r^2.

    Similarly the unit charge will experience a force due to q2 and it is 'F2' = k.16.q2.(1)/9.r^2 ?

    For the point P to not experience any displacement due to the forces 'F1' and 'F2' they have to be equal an opposite.

    k.16.q1.(1)/r^2. = k.16.q2.(1)/9.r^2

    q1/q2 = 1/9.

    Charges are Like that is both are +ve or -ve.

  • E1 = E2

    k * q1/x^2 = k * q2/(r - x)^2

    --> x = (1/4)r

    q1/q2 = (x/r - x)^2

    = (1/4 r/3/4 r)^2

    = (1/3)^

    --> q1/q2 = 1/9

    --> q1 : q2 = 1 : 9

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