The electrostatic force acting on a particle has a magnitude of: The electrostat

Question

The electrostatic force acting on a particle has a magnitude of:

The electrostatic force exerted on a particle by the uniform electric field is directed:

During the period of time it takes the particle to travel a distance of 200 mm parallel to the electric field, the work done on the particle by the electrostatic force is equal to:

The final kinetic energy of the particle is equal to:

During the period of time it takes the particle to travel a distance of 200 mm parallel to the electric field, the change in the electrostatic potential energy of the system of particle plus uniform electric field, the change in the electrostatic potential energy of the system of particle plus uniform electric field is equal to:

The constant acceleration experienced by the particle has a magnitude of:

The constant acceleration experienced by the particle is directed:

How much time did it take the particle to travel the 200 mm parallel to the uniform electric field:

The particle's final speed is equal to:

Please answer all parts of the problem and show all work. Thanks!

A negatively charged particle of mass m = 5.89 x 10-23 g and q =-160 x 1 0-19 C is immersed in a uniform 1.50x 103 NIC electric field pointing in the positive x direction, as shown in the figure. The pa s initially raveling paallel to the uniform electric field at a speed of 1.00x10 m/s. Assume that the gravitational force acting on the particle is negligible compared to the electrostatic force acting on it. Consider a period of time during which the particle travels a distance of 2.00x 102 mm parallel to the uniform electric field. uniform E (1500 N/C) Vi 200 mm fX

Explanation / Answer

here,
mass of particle, m = 5.89 * 10^-23 kg
charge on particle, q = -1.6*10^-19 C
Electric field, E = 1.50*10^3 N/C
Velocity of particle, v = 10^5 m/s
distance travelled between the plates, d = 200 mm = 0.2 m

Part A:
Electrostatic force in electric is given as,
F = q*E
F = -1.6*10^-19 * 1.50*10^3
F = - 2.4*10^-16 N

Part B:
force will act in -x direction as charge is negative.

Part C:
Work donw, w = force * Distance travelled
w = 2.4 * 10^-16 * 0.2
w = 4.8 * 10^-17 J

Part D:
Kinetic Energy of particle, KE
KE = 0.5 * m * v^2
KE = 0.5 * 5.89 * 10^-23 * (10^5)^2
KE = 2.945*10^-13 J

Part E:
Electric potential Energy experied by particle, U
U = -K*q^2/d
U = -9*10^9*(-1.6*10^-19)^2/0.2
U = 1.152*10^-27 J

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