Answer to Question #311257 in Physics for Jay

Question

A uniform electric field of 500 N/C is established between two oppositely charged metal plates. A particle with a charge of +0.003 C is moved from the bottom (negatively charged) plate. The string is tied to the charged that is pulling it upward. The distance between the plates is 5 cm.

a.      What is the change in the potential energy of the charge?

b.      What is the change in electric potential from the bottom to the top plate? 

Answer 1

a.      What is the change in the potential energy of the charge?

$$\Delta E_p=-W=-qEd\cos\theta\\ =-0.003 *500*0.05*(-1)=0.075\:\rm J$$

b.      What is the change in electric potential from the bottom to the top plate? 

$$\Delta V=\frac{\Delta E_p}{q}=\frac{0.075}{0.003}=25\:\rm V$$