Charges q_1 and q_2 exert repulsive forces of 15 N on each other. What is the re
ID: 1567627 • Letter: C
Question
Charges q_1 and q_2 exert repulsive forces of 15 N on each other. What is the repulsive force when their separation is increased so that their final separation is 160% of their initial separation? A) 1.3 N B) 5.9 N C) 9.8 N D) 8.7 N E) 6.4 N An electric field with a magnitude of 6.0 times 10^4 N/C is directed parallel to the positive y axis. A particle with a charge q = 4.8 times 10^-19 C is moving along the x axis with a speed v = 3.0 times 10^6 m/s. The force on the charge is approximately A) 8.6 times 10^-8 N perpendicular to the xy plane. B) 2.9 times 10^-14 N in the y direction. c) 8.6 times 10^-8 N in the x direction. D) zero. E) 2.9 times 10^-14 N in the x direction A solid sphere of radius a is concentric with a hollow sphere of radius b, where b > a. If the solid sphere has a charge +Q and the hollow sphere a charge of -Q, the electric field at radius r, where r > b, is which of the following, in terms of k = (4 pi epsilon_0)^-1 ? A) Kq/r^2 B) 2kQ/r^2 C) Kq/a^2 D) kQ/b^2 E) zeroExplanation / Answer
force between charges is proportional to 1/r2
where r is distance between charges. When final sepration 160% of initial
r' = 1.6 r
force is multiplied by factor of 1/ 1.62
Hence new force is 15/1.62 = 5.9 N
2) Force on a charge in electric field
F = q E
Force is in dircetion of field if charge is +ve and opp to direction of field if charge is -ve
Hence F = 4.8x10-19 * 6x104 = 2.9x10-14 in the direction of field that is y direction.
3) Field created by charge on two shells,are equal in magnitude,that is KQ/r2
but oppsite in direction, as charges are Q and -Q.
hence net field is zero