10. -1 points SerPSE9 25.P.009 A particle having charge q = +1.60 C and mass m =

Question

10. -1 points SerPSE9 25.P.009 A particle having charge q = +1.60 C and mass m = 0.010 0 kg is connected to a string that is L = 1.00 m long and tied to the pivot point P in the figure below. The particle, string, and pivot point all lie on a frictionless, horizontal table. The particle is released from rest when the string makes an angle -60.0 with a uniform electric field of magnitude E = 330 V/m. Determine the speed of the particle when the string is parallel to the electric field My m/s Top view

Explanation / Answer

Given,

q = 1.6 uC ; m = 0.01 kg ; L = 1 m ; theta = 60 deg ; E = 330 V/m

let v be the speed.

The potential at the charge's position is:

V = E d = E L cos(theta)

using conservation of energy

1/2 m v^2 = q E L - q E L cos(theta)

v = sqrt [2 q E L (1 - cos(theta)]/m

v = sqrt {2 x 1.6 x 10^-6 x 300 x 1 [1 - cos(60)]/0.01} = 0.219 m/s

Hence, v = 0.219 m/s (= 0.22 m/s aprox)

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