Voltage drop across a capacitor

In summary, we have a parallel plate capacitor with a positive charge on the left plate and a negative charge on the right plate. The electric field between the plates has a magnitude of 100 kV/m and the plates have an area of 2 × 10-3 m2 and a spacing of 6 × 10-3 m. By using the equations q=vc and PE=QV/2, we can calculate the energy stored in the capacitor to be 2.95133333333333e-012 and 1.7708e-9 respectively. The equation Q=∆V*C may also be used for calculating the voltage drop.
  • #1
zooboodoo
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0

Homework Statement



The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge -Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 × 10-3 m2, and the spacing between the plates is 6 × 10-3 m. There is no dielectric between the plates.

Calculate the energy stored in the capacitor.

Homework Equations


q=vc , PE=c*v^2/2 , PE=Q^2 / 2C , PE=QV / 2



The Attempt at a Solution



I calculated the Capacitance by C=kEoA/Separation = (2.95133333333333e-012)
I also calculated Q by E * Eo *A= Q = 1.7708e-9

I wasn't sure if "Voltage drop" entailed a different equation? I've gone through the powerpoints from my class and have not seen any other equations. My book hopefully comes in the mail soon :-\ any help would be appreciated
 
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  • #2
I wasn't sure if "Voltage drop" entailed a different equation
I believe that would entail Q=∆V*C which you've already got up there :smile:
 
  • #3
Sigh, power of 10 error. It was the right equation, thanks though.
 

FAQ: Voltage drop across a capacitor

1. What is voltage drop across a capacitor?

Voltage drop across a capacitor refers to the decrease in voltage that occurs when a capacitor is placed in a circuit. This decrease in voltage is due to the storage of electrical energy in the capacitor's electric field.

2. How is voltage drop across a capacitor calculated?

The voltage drop across a capacitor can be calculated using the formula V = Q/C, where V is the voltage drop, Q is the charge stored in the capacitor, and C is the capacitance of the capacitor.

3. Why does voltage drop across a capacitor occur?

Voltage drop across a capacitor occurs because a capacitor has the ability to store electrical energy in its electric field. When a capacitor is placed in a circuit, it will charge and discharge, causing a decrease in voltage across it.

4. How does the voltage drop across a capacitor affect the overall circuit?

The voltage drop across a capacitor can affect the overall circuit by causing a time delay in the flow of current. This can be useful in controlling the timing of circuits and can also help to filter out unwanted noise.

5. How can the voltage drop across a capacitor be reduced?

The voltage drop across a capacitor can be reduced by increasing the capacitance of the capacitor or by decreasing the rate of change of the voltage across it. Additionally, using multiple capacitors in parallel can also reduce the voltage drop across each individual capacitor.

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