A particle having charge q = +2.40 A mu C and mass m = 0.0100 kg is connected to

Question

A particle having charge q = +2.40 A mu C and mass m = 0.0100 kg is connected to a string that is L = 1.30 m long and is 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 theta = 60.0 degree with a uniform electric field of magnitude E = 300 V/m. Determine the speed of the particle when the string is parallel to the electric field (point a in the figure).

Explanation / Answer

at Equillibirum point ,

Solving it using conservation of energy

speed Vf^2 = (2q EL (1-cos theta)/m    (formula)

on subtituing

Vf^2 = (2 * 2.4 *10^-6* 300* 1.3 * (1- cos 60)/0.01

Vf = 0.3059 m/s (answer)

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