One point charge of -12.0muC is located at x = 0.0m and a second charge of -24.0

Question

One point charge of -12.0muC is located at x = 0.0m and a second charge of -24.0muC is located at x = 6.0 m. (a) Calculate the point where the electric field would be zero. (b) Calculate the total force on a 10muC charge placed at the point calculated in part (a). (c) Calculate the electric field at a point on the x-axis at x = 12.0 m. (d) Calculated the total force on a 20muC charge placed at the position given in part (c). (e) Calculate the acceleration on the 20muC charge just as it is released from the x = 12.0 m if it has a mass of 5 times 10-3 kg. (Worth )

Explanation / Answer

Given charges are

        q1 = -12*10^-6 C

        q2 = -24*10^-6 C

Seperation between the two charges d = 6 - 0 = 6 m

Let x be distance of the point P from q1 such that at which Net Electric field is Zero.

Electric field due to q1 at P, E1 = Kq1/x^2

                                                  where K is electrostatic constant.

Electric field due to q2 at P, E2 = Kq2/(d - x)^2

For the given problem Net electric field must be zero only when E1 = E2.

So Kq1/x^2 = Kq2/(d - x)^2

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

12/x^2 = 24/(6 - x)^2

1/x^2 = 2/(6 - x)^2

Taking square root on both sides we get

   1/x = 1.414/(6 - x)

    6 - x = 1.414 x

   2.414 x = 6

x = 4.243 m

Hence Net electric field is Zero at a point which is 4.243 m from charge - 12 C(Origin).

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