Effects of capacitors on potential difference

In summary, the potential difference across each capacitor is proportional to the amount of charge that is placed on the capacitor.
  • #1
nagyn
26
0

Homework Statement


Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the quantity of charge you place on capacitor 1 is six times the quantity you place on capacitor 2.

How do the potential differences across each of the two capacitors compare to each other? (V1/V2 = ?)

Homework Equations



I tried two different approaches:

V(A->B) = -W(A->B)/qtest = -(integral of E*dl from A to B)

V = q/C
C = εA/d

The Attempt at a Solution



q1 = 6q2
r1 = 2q2

V1/V2 = (kq1/r1^2)*r1 / (kq2/r2^2)*r^2 = (6q2/2r2)/(q2/r2) = 3

This is not correct.

C1/C2 = d1/d2 = 2

V = q/C = 6/2 = 3

I must be misunderstanding potential difference but I'm not sure where I'm going wrong.
 
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  • #2
? Why do you think 3 is wrong ?
 
  • #3
Cutter Ketch said:
? Why do you think 3 is wrong ?

Is it not wrong? When I tried the answer on MasteringPhysics it was marked as incorrect.
 
  • #4
Well it's pretty straight forward. V is proportional to q and inversely proportional to d, and that is all that changed. Hard to get a different answer.
 
  • #5
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$
 
  • #6
gneill said:
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$

Oh, I suck! I didn't even look at the equations. It just seemed so obvious and natural that the potential difference MUST get bigger as the plates get closer. You know, it seemed so obvious that I'm having trouble getting my head back on straight.
 
  • #7
Sorry about that :smile:
 
  • #8
gneill said:
Capacitance is inversely proportional to d, so potential is proportional to d. That is:
$$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
$$V \propto q \cdot d$$

Wow, I can't believe I messed that up. Thank you!
 

FAQ: Effects of capacitors on potential difference

1. How do capacitors affect potential difference?

Capacitors store electric charge, which can affect the potential difference in a circuit. When a capacitor is fully charged, it creates a potential difference between its plates, leading to a flow of current when the circuit is completed.

2. Do capacitors increase or decrease potential difference?

Capacitors do not inherently increase or decrease potential difference. Instead, they can store potential difference when they are charged and release it when they are discharged.

3. How does the capacitance value affect potential difference?

The capacitance value of a capacitor affects the potential difference it can hold. A higher capacitance value means the capacitor can store more charge, resulting in a higher potential difference.

4. Can capacitors stabilize potential difference in a circuit?

Yes, capacitors can be used to stabilize potential difference in a circuit. This is because they can store and release charge, helping to smooth out fluctuations in potential difference.

5. How do different types of capacitors affect potential difference?

Different types of capacitors have different capacitance values, which can affect the potential difference in a circuit. Additionally, different types of capacitors may also have varying levels of leakage and internal resistance, which can impact potential difference.

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