# Potential Difference of Capacitors

by AdKo
Tags: capacitors, difference, potential
 P: 19 1. The problem statement, all variables and given/known data A parallel-plate capacitor has an area of 5 cm^2, and the plates are separated by 1mm with air between them. The capacitor stores a charge of 400pC. a) What is the potential difference across the plates of the capacitor? b) What is the magnitude of the uniform electric field in the region between the plates? 2. Relevant equations 1. $$U_{c}=\frac{1}{2}C(\Delta V)^2$$ 2. $$C=\epsilon_{0}\frac{A}{d}$$ 3. The attempt at a solution a) Using equation #1, $$\Delta V=\sqrt{\frac{2(400*10^-12C)}{(4.425*10^-10F)}}$$ V=1.3446 V ... however, the answer in the back of the book is: 90.4 V What am I doing wrong here? I'm pretty sure I'm using the right equation. b) E=V/d I plugged in the book's answer for V and used the given distance to find the electric field magnitude. Help me on part a please? Thanks!
 HW Helper P: 1,542 Your first equation listed doesn't make sense for this problem. That is the equation for the energy stored in a capacitor, which you don't know (U does not equal the charge). Try to find a simplier equation containing the three things you know. For your calculation of C using equation 2: You answer isn't quite right. I suspect you didn't convert the area from cm^2 to m^2 properly.
P: 19
 Quote by hage567 Your first equation listed doesn't make sense for this problem. That is the equation for the energy stored in a capacitor, which you don't know (U does not equal the charge). Try to find a simplier equation containing the three things you know. For your calculation of C using equation 2: You answer isn't quite right. I suspect you didn't convert the area from cm^2 to m^2 properly.
thanks for all your help. I solved by using C=Q/V that was easy.

 Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 1 Introductory Physics Homework 2 Electrical Engineering 5 Introductory Physics Homework 1