Two point charges, q1 and q2, are placed 0.30 m apart on the x- axis, as shown i

Question

Two point charges, q1 and q2, are placed 0.30 m apart on the x- axis, as shown in the figure. Charge q1 has a value of -3.0 x 10^-9 C. The net electric field at point P is zero.

A) Why is the sign of charge q2 positive? explain?

B) Calculate the magnitude of charge q2.

C) How do the magnitude and direction of the electrostatic force on q1 compare to the magnitude and direction of the electrostatic force on q2 explain?

D) Calculate the magnitude of the electrostatic force on q2 and indicate its direction.

E) Determine the x-coordinate of the point on the line between the two charges at which the electric potential is zero.

F) How much work must be done by an external force to bring and electron from infinity to the point at which the electric potential is zero? explain why.

Explanation / Answer

A) E1 = E2 (magnitudes are same, but direction is opposite)

k*q1/0.1^2 = k*q2/0.4^2

q2 = (0.4/0.1)^2 *q1 = +48*10^-6 C

q2 should be positive.
then E2 is in -x direction and E1 is in +x direction.
Enet = 0

c)on q1 force acts +x axis

on q2 force acts towards -x axis.

these two forces are equal in magnitude.
F = k*q1*q2/0.3^2

D)F = k*q1*q2/0.3^2 = 14.4 N towards -x axis

E) k*q1/x = k*q2/0.3-x

solving above equation we get x = 0.0176 m from left charge

the x coardinate of this point = -0.1+0.0176 = -0.0824 m

F) Zero.

W = q*(v1-v2)
here v1 = v2 = 0
so no need to do any work

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