Question

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A. The particle subsequently passes through point B which has an electric potential of +1.5 kV relative to point A. Determine the kinetic energy of the particle as it moves through point B.

Answer 1

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

Where:

K: is the kinetic energy

$E_{p}$: is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV                                    

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!