A charged particle (q = -8.0 mC), which moves in a regionwhere the only force ac

Question

A charged particle (q = -8.0 mC), which moves in a regionwhere the only force acting on the particle is an electric force,is released from rest at point A. At point B the kinetic energy ofthe particle is equal to 4.8 J. What is the electric potentialdifference V?
please show steps A charged particle (q = -8.0 mC), which moves in a regionwhere the only force acting on the particle is an electric force,is released from rest at point A. At point B the kinetic energy ofthe particle is equal to 4.8 J. What is the electric potentialdifference V?
please show steps

Explanation / Answer

Charge q = -8 * 10 ^ -3 C

Kinetic energy at point A is K = 0

Kinetic energy at point B is K ‘ = 4.8 J

Chang ein kinetic energy K ‘ – K = 4.8 J

This is equal to potential energy

i.e., Vq = K ‘-K

from this potential difference V = ( K ‘-K ) / q

                                                    = 600 Volt

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