(III) Suppose two thin flat plates measure 1.0 m ×1.0 m...

(III) Suppose two thin flat plates measure $1.0 \mathrm{~m} \times 1.0 \mathrm{~m}$ and are separated by $5.0 \mathrm{~mm}$. They are oppositely charged with $\pm 15 \mu \mathrm{C}$. (a) Estimate the total force exerted by one plate on the other (ignore edge effects). (b) How much work would be required to move the plates from $5.0 \mathrm{~mm}$ apart to $1.00 \mathrm{~cm}$ apart?

Key Concepts

Work Done in Changing a Capacitor Configuration

The work involved in moving charged plates is directly related to the change in the system’s stored electrostatic energy. As the distance between the plates changes, the capacitance and the energy stored in the electric field also change. This concept is useful for analyzing how much work is required to modify the separation between the plates in a capacitor setup, translating the physical displacement into energy considerations.

Electrostatic Energy and Capacitance

In a system of parallel plates, the energy stored in the system is also a key concept. Capacitors store electrical energy by means of the separation of charge between the plates. The capacitance, which depends on the plate area, separation distance, and the permittivity of the medium between the plates, is crucial for calculating the electrostatic potential energy. Understanding this helps in quantifying the amount of work required to change the configuration of the capacitor.

Electrostatic Force on a Charged Object

This concept refers to the force experienced by a charged object when placed in an external electric field. The force can be understood as the product of the charge and the electric field acting on it. In the context of parallel plates, it is used to evaluate the mutual attractive force that one plate exerts on the other by considering the field due to a single plate acting on the charge distribution of the opposite plate.

Superposition Principle in Electrostatics

The superposition principle states that when there are multiple sources of electric fields, the net field at any point is the vector sum of the fields produced by each individual charge or distribution. This is essential when analyzing the field between two oppositely charged plates, as it allows the calculation of the effective field by summing the contributions from both plates.

Uniformly Charged Plate Field

This concept involves the derivation of the electric field created by an infinite or sufficiently large flat plate with a uniform charge distribution. The field is constant in magnitude and direction close to the surface of the plate, and it is calculated using Gauss's law. Understanding this helps in determining the contribution of each charged plate to the overall electric field in configurations such as parallel plate capacitors.

Transcript

00:01 Okay, so we're doing chapter 22 problem 45.

00:04 So this problem says we have two thin flat plates of one meter by one meter.

00:11 And they're separated by a distance of five millimeters.

00:16 And they are oppositely charged with plus or minus 15 microculems on me.

00:22 So if this is their total charge, that means they're going to have charge densities of plus or minus 15 microtulums on me.

00:32 Microculems per meter squared since the area is just one meter squared.

00:37 Makes it nice and simple for us.

00:40 So part a says estimate the total force exerted on one plate from the other.

00:46 So let's go ahead and define some things here.

00:56 So if we find the force on the negative plate from the positive plate, so let's call plate b.

01:09 The negative plate and plate a, the positive plate, so that if we're looking the force on plate b, this is given as the charge on plate b, times the electric field produced from plate a.

01:28 So if we look at the total, or look at example 227, we can see that the electric field from a plate so let's rewrite this equation as charge on b times sigma a over to epsilon not.

01:50 Well, the total charge on a is just sigma b times the area and then times this.

01:59 So this is what the force is.

02:04 So we can now simplify this as 15 microculems per meter squared times one meter squared times negative.

02:14 This one's negative...

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