Capacitance calculation

In summary, to find the capacitance of a parallel-plate capacitor with Earth and a cloud layer 700m above, you need to convert the area to square meters and use the correct value for the permittivity of free space. Make sure to also pay attention to units, significant figures, and use the correct formula for capacitance.
  • #1
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Homework Statement


Consider Earth and a cloud layer 700m above Earth to be the plates of a parallel-plate capacitor. If the cloud layer has an area of 1.9 km^2, what is the capacitance?

Homework Equations


Q = CV
V = Ed
E = Q/epsiolon*A

The Attempt at a Solution


Solving for C gives us C=Q/V. Substituting Ed for V gives us C=Q/Ed. Substituting Q/epsilon*A gives us C = epsilon*A/d. Now I just plug in numbers and I should get the value for C, right? I'm putting in the values given but I am not getting the right answer...any suggestions? Thanks in advance to anyone who helps.
 
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  • #2


it is important to always double check your calculations and make sure you are using the correct units. In this case, the area given is in square kilometers, but the equation for capacitance requires the area to be in square meters. So, the first step would be to convert the area from 1.9 km^2 to 1.9 x 10^6 m^2.

Next, make sure you are using the correct value for the permittivity of free space (epsilon). It is typically denoted by the symbol ε0 and has a value of approximately 8.85 x 10^-12 C^2/Nm^2.

Once you have the correct units and values, you can plug them into the equation C = ε0A/d and solve for the capacitance. Remember to also pay attention to significant figures and round your answer to the appropriate number of decimal places.

If you are still not getting the correct answer, double check your calculations and make sure you are using the correct formula for capacitance. It is also helpful to show your work and any assumptions you made in your solution, as this can help identify any errors.
 
  • #3


I would suggest double checking your calculations and units to ensure accuracy. Additionally, make sure you are using the correct value for the permittivity of free space (epsilon) and that all units are consistent (e.g. converting 700m to km). If you are still not getting the correct answer, it may be helpful to show your work and calculations for further assistance.
 

What is capacitance and why is it important in circuit design?

Capacitance is the ability of a conductor to store electrical charge. It is important in circuit design because it determines how much charge can be stored and how quickly it can be released, which affects the performance of the circuit.

How do you calculate the capacitance of a capacitor?

The capacitance of a capacitor can be calculated by dividing the amount of charge stored on one plate by the potential difference between the plates. This can be represented by the formula C = Q/V, where C is capacitance, Q is charge, and V is potential difference.

What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by the distance between the plates, the area of the plates, and the type of material between the plates (known as the dielectric). A closer distance, larger plate area, and higher dielectric constant will result in a higher capacitance.

How do you calculate the total capacitance in a circuit with multiple capacitors?

To calculate the total capacitance in a circuit with multiple capacitors, you can use the formula C = C1 + C2 + C3 + ..., where C is the total capacitance and C1, C2, C3, etc. are the individual capacitances of each capacitor in the circuit.

What is the difference between series and parallel capacitors and how does it affect the total capacitance?

In a series circuit, the capacitors are connected end-to-end, and the total capacitance is less than the individual capacitances. In a parallel circuit, the capacitors are connected side-by-side, and the total capacitance is equal to the sum of the individual capacitances. This is because in a series circuit, the effective distance between the plates increases, while in a parallel circuit, the effective plate area increases.

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