# Stuck on the last problem.

1. Feb 24, 2004

### NINHARDCOREFAN

https://hw.utexas.edu/tmp/Muddam1/1077653690Xuj.pdf [Broken]

I did this: I manipulated $$E = \frac{q}{\epsilon*A}$$ to get $$q = E*\epsilon*A$$ (E= electric strength), so 1.6e7*8.85e-12*.0246*.0501 = 1.74516e-7 but because the HW asks the answer to the 1e-6, I divided 1.74516e-7/1e-6= .174516 but it's wrong does anyone know why?

Last edited by a moderator: May 1, 2017
2. Feb 24, 2004

### Aerospace

Your link does not work.

3. Feb 24, 2004

### NINHARDCOREFAN

Whoa....

Ok this is the problem

009 (part 1 of 2) 10 points
A parallel-plate capacitor of dimensions
2:46 cm £ 5:01 cm is separated by a 0:61 mm
thickness of paper.
Find the capacitance of this device. The
dielectric constant · for paper is 3.7. Answer
in units of pF.

I got this right.

010 (part 2 of 2) 10 points
What is the maximum charge that can be
placed on the capacitor? The electric strength
of paper is 1:6 £ 107 V=m. Answer in units of 1e-6c.
this is the one I need help on

4. Feb 24, 2004

### NINHARDCOREFAN

...

Never mind, I got it.

5. Feb 24, 2004

### Aerospace

So what's the answer?

6. Feb 24, 2004

### NINHARDCOREFAN

I used a different formula to get the answer: $$q = E*k\epsilon*A$$

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