18.1) Two point charges are fixed in place on an x axis. Particle 1 of charge q1

Question

18.1) Two point charges are fixed in place on an x axis. Particle 1 of charge q1 = -4q is located at x = +d, while particle 2 of charge q2 = q is located at x = 0. Consider a point P located on the x axis at x = +2d. Let q = 12 microcoulombs and d = 0.3 m.

a) What is the direction of the electric field at point P?

b) What is the magnitude of the electric field at point P?

c) A 2-microcoulomb charge is placed at point P and then released from rest. What is the electric force on this charge immediately after it is released?

Explanation / Answer

here,

taking direction + ve in +x direction

q1 = - 4 q is at x = d

q2 = q is at x = 0

p = 2 d

electric feild , E = E2 - E1

E = k * q2/(2*d)^2 - k * q1 /d^2

E = 9 * 10^9 * 12 * 10^-6 /( 2 * 0.3)^2 - 9 * 10^9 * 4 * 12 * 10^-6 /( 0.3)^2

E = - 4.5 * 10^6 N/C

a)

the direction of electric feild is towards -x axis

b) the magnitude of the electric feild is -4.5 * 10^6 N/C

c)

q' = 2 * 10^-6 C

the electric force on charge , F = q' * E

F = - 2 * 10^-6 * 4.5 * 10^6 N

F = - 9 N

the electric force on this charge is 9 N in -ve x direction

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