How to Calculate Capacitance for Two Slanted Conducting Plates?

The capacitance of two plates facing each other at an angle can be determined using the equation C = (εA)/d, where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates. In this problem, plate A and B have dimensions of 10x20 cm and 10x(20sec30) (angles in degrees) respectively. By taking one strip of the plate with width dx and length L and finding the integration, the capacitance can be calculated. The variable d' represents the variance distance of the slanted plate from the horizontal line, which can be found using the given values in the problem.
  • #1
blackblanx
16
0

Homework Statement



Determine the capacitance of two conducting plates facing each other at an angle as shown (in link) . Plate A and B have dimensions of 10x20 cm and 10x (20sec30) (angles in degrees)

Homework Equations


C=( [tex]\epsilon[/tex] A)/d


The Attempt at a Solution


not quite sure can someone help me get the ball rolling or the floor moving
 

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  • #2
If you take one strip of the plate of width dx and length L.
The capacity of that strip is given by
ΔC = εο*L*dx/(d + d')
From the figure write down d' in terms of x and θ.
Find the integration to get the capacitance C.
 
  • #3
What are you using for d' ?
 
  • #4
blackblanx said:
What are you using for d' ?
d' is the variance distance of the slanted plate from the horizontal line.
In the problem, numerical values are not correct.
One end of the slanted plate is 1 mm from the top plate,. And the other end is 0.1 mm from the top plate. So tan θ = 0.9 mm/0.1 mm which is not equal to tan(30)
 

Related to How to Calculate Capacitance for Two Slanted Conducting Plates?

1. How do you find capacitance using calculus?

To find capacitance using calculus, you need to have the formula for capacitance, which is C=Q/V, where C is capacitance, Q is charge, and V is voltage. You also need to have a function that describes the voltage across the capacitor as a function of time. Then, you can use integration to find the total charge and voltage over a specific time interval, and plug those values into the capacitance formula to find the capacitance.

2. What is the difference between finding capacitance using calculus and using other methods?

The main difference is that using calculus allows for finding the capacitance of more complex systems, where the voltage across the capacitor is not constant. Additionally, using calculus can provide more precise results compared to other methods, which may rely on approximations or simplifications.

3. Can calculus be used to find capacitance for any type of capacitor?

Yes, calculus can be used to find capacitance for any type of capacitor, as long as the voltage and charge can be expressed as functions of time. This includes capacitors with varying geometries, dielectric materials, and circuit configurations.

4. Are there any limitations to using calculus for finding capacitance?

One limitation is that the function describing the voltage across the capacitor must be known or able to be determined. This may not always be possible for certain systems, making it difficult to use calculus to find capacitance. Additionally, calculus may be more time-consuming and complex compared to other methods, especially for simpler capacitor systems.

5. How can finding capacitance using calculus be applied in real-world scenarios?

Finding capacitance using calculus is essential in designing and analyzing electronic circuits and systems. It allows for accurate prediction of the behavior of capacitors and their interactions with other components. This information is crucial in fields such as electrical engineering, physics, and other sciences that utilize capacitors in their research and applications.

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