Potential Difference of Capacitors

AI Thread Summary
The discussion centers on calculating the potential difference and electric field of a parallel-plate capacitor with given parameters. The initial attempt at finding the potential difference using the wrong equation led to confusion, as the user misapplied the energy stored formula instead of using the charge-capacitance relationship. The correct approach involves using the formula C=Q/V, which simplifies the calculation. Additionally, there was an issue with converting the area from cm² to m², which affected the capacitance calculation. Ultimately, the user successfully resolved the problem by applying the correct equations.
AdKo
Messages
19
Reaction score
0

Homework Statement


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?


Homework Equations


1. U_{c}=\frac{1}{2}C(\Delta V)^2
2. C=\epsilon_{0}\frac{A}{d}

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? :smile: Thanks!
 
Last edited:
Physics news on Phys.org
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.
 
hage567 said:
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 :smile: that was easy. :biggrin:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top