Electric Potential Energy — Question 6

Question 6

Gravity and electric work can oppose each other during a fall

A particle of mass m and positive charge q is released from rest and falls a vertical distance h in a uniform upward electric field E. Choose upward coordinate y and assume $qE=.600mg$. Ignore drag.

1. Find gravitational work, electric work and the net kinetic-energy gain.

2. Find the final speed and the field strength that would instead make the particle remain at rest in ideal force balance.

Question 6 - Solution

Gravity and electric work can oppose each other during a fall

(a) Method and calculation

$$W_g=mgh,\quad W_e=qE(-h)=-.600mgh.$$$$\Delta K=W_g+W_e=.400mgh,\quad\Delta U_{\mathrm{total}}=-.400mgh.$$

(b) Interpretation and further calculation

$$\tfrac12mv_f^2=.400mgh\Rightarrow v_f=\sqrt{.800gh}.$$$$qE_{\mathrm{balance}}=mg\Rightarrow E_{\mathrm{balance}}=mg/q.$$

Physical check

The electric potential energy increases during this downward motion, while the larger gravitational decrease supplies the kinetic energy.