Question

What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

Answer 1

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, $U_{electric,b} - U_{electric,a}$ as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2×$10^{-6}$ J

(b) KE = 2×$10^{-6}$ J

Explanation: Potential Energy (U) is the amount of work done due to its position or condition and its unit is Joule (J). Kinetic Energy (KE) is the ability to do work by virtue of velocity and the unit is also (J). Mechanical Energy is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2×$10^{3}$ N/C

q = 2×$10^{-9}$ C

d = 50 - (-30) = 80×$10^{-2}$ = 8×$10^{-1}$m

ΔU = $U_{electric,b} - U_{electric,a}$ = Eqd

$U_{electric,b} - U_{electric,a}$ = 2×$10^{3}$ .  2×$10^{-9}$ . 8×$10^{-1}$

ΔU = 3.2×$10^{-6}$ J

(b) Mechanical Energy is constant, so:

$KE_{i} + U_{i} = KE_{f} + U_{f}$

Since the initial position is zero and there is no initial kinetic energy:

$KE_{f} = - U{f}$

$KE_{f} =$ - (2×$10^{3}$. 2×$10^{-9}$ . 5×$10^{-1}$)

$KE_{f} = - 2.10^{-6}$ J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.