Capacitance problem (verify answer)

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In summary, the capacitance of a parallel plate capacitor with A=0.04 m^2, distance of separation between plates = 2mm, relative permittivity of dielectric = 6, and permittivity of free space = 8.854x10^-12 is 1062.48 pF.
  • #1
jayjay112
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Find the capacitance of a parrell plate capacitor in pF with A=0.04 m^2
distance of separation between plates = 2mm relative permittivity of dielectric = 6
and permittivity of free space = 8.854x10^-12


C = (6)(8.854x10^-12)(0.04) / 0.002

= 1.06248x10^-9 F

in pF = 1062.48 pF

Thanks
 
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  • #2
jayjay112 said:
Find the capacitance of a parrell plate capacitor in pF with A=0.04 m^2
distance of separation between plates = 2mm relative permittivity of dielectric = 6
and permittivity of free space = 8.854x10^-12


C = (6)(8.854x10^-12)(0.04) / 0.002

= 1.06248x10^-9 F

in pF = 1062.48 pF

Thanks

You had it until you wrote the last answer.

1.062 pF
 
  • #3
LowlyPion said:
You had it until you wrote the last answer.

1.062 pF

But the answer 1.062x10^-9 isn't the same as 1.062x10^ -12 or 1.062 pF
 
  • #4
jayjay112 said:
But the answer 1.062x10^-9 isn't the same as 1.062x10^ -12 or 1.062 pF

But 1062 * 10-12 = 1.062 * 10-9

10-12 is pico, so sorry you are right. I confused nano-
 
  • #5
LowlyPion said:
But 1062 * 10-12 = 1.062 * 10-9

10-9 is pico.


pico is 10^ -12

see link http://en.wikipedia.org/wiki/Pico-

Thanks for your help with verifing my results.
 
  • #7
LowlyPion said:
But 1062 * 10-12 = 1.062 * 10-9

10-12 is pico, so sorry you are right. I confused nano-

no problem thanks again
 

Related to Capacitance problem (verify answer)

1. What is capacitance?

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

2. How do you calculate capacitance?

Capacitance is calculated by dividing the charge (in Coulombs) by the potential difference (in Volts) between two conductors. In equation form, it is expressed as C = Q/V.

3. What is the relationship between capacitance and voltage?

The relationship between capacitance and voltage is inverse. This means that as capacitance increases, voltage decreases, and vice versa. In equation form, it is expressed as C = Q/V.

4. How is capacitance affected by the distance between conductors?

The closer the conductors are to each other, the higher the capacitance will be. This is because the electric field between the conductors is stronger, allowing for more charge to be stored.

5. How can you verify the answer to a capacitance problem?

To verify the answer to a capacitance problem, you can use a multimeter or an oscilloscope to measure the voltage and charge in the circuit. You can also use the equation C = Q/V to check if your calculated capacitance is consistent with the measured values. Additionally, you can compare your solution to a known solution or use simulation software to confirm your answer.

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