A positive point charge q_1 creates an electric field of magnitude E_1 at a spot

Question

A positive point charge q_1 creates an electric field of magnitude E_1 at a spot located at a distance r_1 from the charge. The charge is replaced by another positive point charge q_2, which creates a field of magnitude E_2 = E_1 at a distance of r_2 = 2r_1. How is q_2 related to q_1? A 1.4 - mu C point charge is placed between the plates of a parallel plate capacitor. The charge experiences a force of 0.64 N. What is the magnitude sigma of the charge density on either plate of the capacitor? In a vacuum, two particles have charges of q_1 andq_2, where q_1 = + 4.7^mu C. They are separated by a distance of 0.30 m, and particle 1 experiences an attractive force of 4.4 N. What is the value of q_2 with its sign? Two charges attract each other with a force of 8.9 N. What will be the force if the distance between them is reduced to one-ninth of its original value? Two point charges are fixed on the y axis: a negative point charge q_1 = - 35 mu C at v1 = + 0 19 m and a positive point charge q_2 at y2 = + 0.39.m. A third point charge q = + 8.1 mu C is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 28 N and points in the + y direction. Determine the magnitude of q_2.

Explanation / Answer

E = k q / r^2

E2 = E1 => k q2 / r2 = k q1 / r1

q2 / 2 r1 = q1 / r1

q2 = 2 q1 .........Ans

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F = q E

E = (0.64) / (1.4 x 10^-6) = 4.57 x 10^5 N/C

E = sigma / e0

sigam = 4.57 x 10^5 x 8.854 x10^-12

= 4.05 x 10^-6 C/m^2

............................................

F = k q1 q2 / r^2

4.4 = (9 x 10^9) (4.7 x 10^-6)(q2) / 0.30^2

q2 = 9.36 x 10^-6

force is attractive hence q1 and q2 will be of opposite sign.

hence q2 = - 9.36 uC

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