Two point charges are fixed in placed on an x axis. Particle 1 of charge q_1 = q

Question

Two point charges are fixed in placed on an x axis. Particle 1 of charge q_1 = q is located at x = 0, while particle 2 of charge q_2 = -q/2 is located at x= +d. Consider a point P located on the x axis at x = (2/3)d. Let q = 8 nC and d = 0.03 m. What is the direction of the electric field at point P? (+x) What is the magnitude of the electric field at point P? (5.4 times 10^5 N/C). A 1-nC charge is placed at point P and then released from rest. What is the electric force on this charge immediately after it is released? (5.4 times 10^-4 N)

Explanation / Answer

a)The direction of field will be from q1 to q2, means towards right.

b)The field at the point P will be the sum of individual fields of the charges:

Ep = kq1/r1^2 + kq2/r2^2

Ep = 9 x 10^9 x 10^-9 (8/0.02^2 + 4/0.01^2) = 5.4 x 10^5 N/C

Hence, Ep = 5.4 x 10^5 N/C

c)We know that, force is given by:

F =qE

F = 1 x 10^-9 x 5.4 x 10^5 = 5.4 x 10^-4 N

Hence, F = 5.4 x 10^-4 N

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