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:
[tex]\Delta K = \Delta E_{p} = q\Delta V[/tex]
Where:
K: is the kinetic energy
[tex]E_{p}[/tex]: is the potential energy
q: is the particle's charge = 0.8 mC
ΔV: is the electric potential = 1.5 kV
[tex]\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J[/tex]
Now, the kinetic energy of the particle as it moves through point B is:
[tex] \Delta K = K_{f} - K_{i} [/tex]
[tex] K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J [/tex]
Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.
I hope it helps you!
