A parallel-plate air capacitor has a capacitance of 920 pF . The charge on each plate is 2.55μC

A parallel-plate air capacitor has a capacitance of 920 pF ....

A parallel-plate air capacitor has a capacitance of 920 $\mathrm{pF}$ . The charge on each plate is 2.55$\mu \mathrm{C}$ . (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

Key Concepts

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit potential difference. In a parallel-plate capacitor, it depends on the area of the plates, the separation between them, and the dielectric constant of the medium between the plates. The formula C = ??A/d for an air capacitor illustrates how the capacitance decreases when the separation is increased while keeping the plate area constant.

Potential Difference

The potential difference across a capacitor is directly related to the charge stored and inversely related to its capacitance, following the relation V = Q/C. When the capacitance is changed, such as by altering the distance between the plates, the voltage across the capacitor will adjust accordingly, especially under conditions where the stored charge remains constant.

Energy Stored in a Capacitor

The energy stored in a capacitor is given by the formula U = Q²/(2C) when the charge is constant. This energy expression is crucial for understanding how modifications in the physical parameters of the capacitor, such as plate separation, affect the stored energy, thereby influencing the work that must be done during such changes.

Work in Changing Capacitor Geometry

When the geometry of a capacitor, such as the separation between the plates, is altered under constant-charge conditions, the work done on the system is associated with the change in stored energy. Specifically, doubling the separation changes the capacitance and thus the energy stored, and the work required is equal to the difference in these energy levels.

Constant Charge Process

In a constant charge process, such as disconnecting the capacitor from any voltage source, the charge on the plates remains fixed while changes in the capacitor’s configuration alter the voltage and stored energy. This concept is key for determining how modifications in geometry impact the overall behavior of the capacitor, including the resulting potential difference and work done.

Transcript

00:01 Okay, so this problem, we start with a parallel plate capacitor, and it has a 920 picofarad capacitance.

00:17 And the charge is 2 .55 microculum on each plate.

00:33 And first we want to get the potential difference.

00:37 So this is for a, b equals question mark.

00:40 So c is q over v, so v is q over c.

00:43 So it's q over c.

00:45 So i'm going to go ahead and plug those into a calculator.

00:50 So let's do 2 .55 times 10 to the minus 6 divided by 920 times 10 to the minus 12.

01:04 And then i got 2 ,770 volts.

01:18 And next we are concerned with what happens to the potential difference if we double the separation.

01:29 So we want to get v equals question mark, q is the same, and then the distance goes to double.

01:41 So to figure out what happens, i mean, again, we want to evaluate v as q over c.

01:47 The problem tells us that q stays the same.

01:50 So c is going to end up changing when you increase the distance.

01:54 Because of c is equal to epsilon not a over d, if you double the distance, then you're going to half the capacitance.

02:04 And if you half the capacitance, then you'll double the voltage.

02:10 So v goes to 2b.

02:13 And so b then it's just going to be double 2 ,770.

02:20 And so i'll multiply that by two, and i get 5 ,540 volts.

02:32 Another way you could reason this is that the electric field stays the same if you have the same amount of charge.

02:40 And then voltage is the integral of electric field over distance.

02:45 So if you've increased the distance, you've increased the voltage.

02:49 And so then next, we want to know how much work is required to do this.

02:54 Let me box this off.

02:59 So here's a, b, and then let's do c...

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