Question

A charged particle (q = −8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference VB − VA?

Answer 1

Answer:

$V_B - V_A = 600 Volts$

Explanation:

As we know that charge is released from rest and then at point B it is moving with kinetic energy K = 4.8 J

so here by mechanical energy conservation law we can say that gain in kinetic energy of the charge is equal to the loss in electrostatic potential energy.

so here we can say

$U_B + KE = U_A$

$KE = U_A - U_B$

$4.8 = (-8 mC)(V_A - V_B)$

$4.8 = (8 \times 10^{-3})(V_B - V_A)$

$V_B - V_A = 600 Volts$

Answer 2

The electric potential difference of the given particle is; 600 V

What is the Potential Difference?

We are told that the charge is released from rest and then at point B it is moving with a kinetic energy of 4.8 J.

From law of conservation of energy, we can say that gain in kinetic energy of the charge is equal to the loss in electrostatic potential energy.

Thus;

K.E = qVA - qVB

K.E = q(VA - VB)

Where q = -8mC = -8 * 10⁻³ C

4.8 = (-8 * 10⁻³)(VA - VB)

VB - VA = 4.8/(-8 * 10⁻³)

VB - VA = 600 V

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