Calculate capacitance of conductors

In summary, the capacitance of the 2 conductors increases by 10^-10 Farads when the voltage difference between the 2 increases to 250 V.
  • #1
s-f
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Problem: (a) What is the capacitance of the 2 conductors +20nC and -20nC? (b) If the voltage difference between the 2 conductors increase to 250 V, then what else changes?

a). I calculated that ΔV = 100 V. So then I plugged into the formula C = Q/ΔV. I thought 20 nC = 2 x 10^-8 C, which divided by 100 V = 2 x 10^-9C but the answer is 2 x 10^-10 C. What am I doing wrong? I know this is very simple but I have never studied physics before and am terrible at math.

b). charge of the conductors increases to keep capacitance constant, since capacitance is a property of area and distance between plates (I think this is right but please tell me if it's wrong!)

Thanks in advance!
 
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  • #2
s-f said:
Problem: (a) What is the capacitance of the 2 conductors +20nC and -20nC? (b) If the voltage difference between the 2 conductors increase to 250 V, then what else changes?

a). I calculated that ΔV = 100 V. So then I plugged into the formula C = Q/ΔV. I thought 20 nC = 2 x 10^-8 C, which divided by 100 V = 2 x 10^-9C but the answer is 2 x 10^-10 C. What am I doing wrong? I know this is very simple but I have never studied physics before and am terrible at math.

b). charge of the conductors increases to keep capacitance constant, since capacitance is a property of area and distance between plates (I think this is right but please tell me if it's wrong!)

Thanks in advance!

Welcome to the PF.

Where did the 100V come from? Is it given for part a)?
 
  • #3
Thanks. Actually the problem had more steps before this and Ohm's Law was used to calculate Voltage = 100 V (I have the answers so I know it's correct). The part I'm unsure of is how C = 2 x 10^-10 Farads (not Coulombs - I accidentally wrote C instead of F earlier!).
 
  • #4
Anyone? I would really appreciate it : )
 
  • #5


(a) The capacitance of the conductors can be calculated using the formula C = Q/ΔV, where C is the capacitance, Q is the charge, and ΔV is the voltage difference. In this case, the charge is 20 nC for each conductor, which is equivalent to 2 x 10^-8 C. The voltage difference between the conductors is 100 V. Plugging these values into the formula, we get C = 2 x 10^-8 C/100 V = 2 x 10^-10 F.

It seems that you may have made a mistake in converting the charge from nanocoulombs (nC) to coulombs (C). Remember, 1 nC = 1 x 10^-9 C, so the charge for each conductor should be 2 x 10^-8 C, not 2 x 10^-9 C.

(b) You are correct in saying that the charge of the conductors will increase to keep the capacitance constant. This is because capacitance is directly proportional to charge, so as the voltage difference increases, the charge must also increase to maintain the same capacitance. Additionally, the electric field between the conductors will also increase, as it is directly proportional to the voltage difference. This means that the force between the conductors will also increase.
 

1. What is capacitance?

Capacitance is the ability of a conductor to store electrical energy in the form of an electric charge. It is measured in units of Farads (F).

2. How do you calculate capacitance?

Capacitance can be calculated by dividing the amount of electric charge stored on a conductor by the potential difference (voltage) across the conductor. It can also be calculated by multiplying the permittivity of the material by the surface area of the conductor and dividing by the distance between the plates.

3. What factors affect the capacitance of a conductor?

The capacitance of a conductor is affected by the surface area of the conductor, the distance between the plates, and the permittivity of the material between the plates. It is also affected by the type of material used and the presence of any dielectric material between the plates.

4. How can capacitance be increased?

Capacitance can be increased by increasing the surface area of the conductor, decreasing the distance between the plates, and using a material with a higher permittivity between the plates. The use of a dielectric material between the plates can also increase capacitance.

5. Why is capacitance important in electronics?

Capacitance is important in electronics because it allows for the storage of electrical energy and can be used in various circuits for filtering, energy storage, and signal processing. It also plays a crucial role in the functioning of capacitors, which are essential components in many electronic devices.

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