# Application of Gauss' Laws: Large area parallel plate capacitor

• knowlewj01
In summary, the electric field strength inside a parallel plate capacitor with equal but opposite charges is given by E = q/(A * epsilon_0), where q is the charge and A is the area. Doubling the positive charge will result in a doubling of the electric field strength, while changing the negative charge will have the same effect. The percent change in electric field strength can be calculated as (new field strength - original field strength)/(original field strength) * 100%.
knowlewj01

## Homework Statement

The plates of a large area parallel plate capacitor of area A are separated by a short
distance. The plates carry an equal but opposite charge $\pm q$.
(a) What is the electric field strength $E(q,A)$ inside the capacitor?
(b) By how many percent does the electric field strength change if the positive charge is doubled?

## Homework Equations

$\oint\bf{E.n} dS = 4\pi k Q$

Electric Field inside a capacator: $E = \frac{Q}{A \epsilon_0}$

## The Attempt at a Solution

the electric field inside a parallel plate capacator is uniform so it does not matter where we do the surface integral $\oint\bf{E.n} dS$
i choose to do the integral on a plane equal in area to the area of the capacators, i will call this area A.

(a)
$\oint\bf{E.n} dS = 4\pi k Q = \frac{q}{\epsilon_0}$

$E A = \frac{q}{\epsilon_0}$

$E = \frac{q}{A \epsilon_0}$

this is in agreement with the given formula i found.

(b) double the positive charge:

$\oint\bf{E.n} dS = 4\pi k Q = \frac{2q}{\epsilon_0}$

$E = \frac{2q}{A \epsilon_0}$

so this is twice that of in the first case. is this the right way to look at this? i thought what would be the difference if i changed the negative one instead, since in this solution i am ignoring the fact that there is a negative charge at all. i think that it would have the same effect.

the percent change can be worked out as:percent change = (\frac{2q}{A \epsilon_0} - \frac{q}{A \epsilon_0})/ \frac{q}{A \epsilon_0} * 100percent change = 100%

## 1. What is Gauss' Law and how does it apply to large area parallel plate capacitors?

Gauss' Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed within that surface. In the context of large area parallel plate capacitors, Gauss' Law can be used to calculate the electric field between the plates by considering the flux through a Gaussian surface that encloses the plates.

## 2. How does the electric field between the plates of a large area parallel plate capacitor vary with distance?

The electric field between the plates of a large area parallel plate capacitor is constant and does not vary with distance. This is because the plates are infinitely large and the electric field lines are parallel and uniform between them.

## 3. What is the capacitance of a large area parallel plate capacitor?

The capacitance of a large area parallel plate capacitor is directly proportional to the plate area and inversely proportional to the distance between the plates. It can be calculated by dividing the electric permittivity of the medium between the plates by the distance between the plates.

## 4. How does the capacitance of a large area parallel plate capacitor change if the distance between the plates is increased?

If the distance between the plates of a large area parallel plate capacitor is increased, the capacitance decreases. This is because the electric field between the plates weakens, resulting in a decrease in the amount of charge that can be stored on the plates.

## 5. How can Gauss' Law be used to determine the electric field between the plates of a large area parallel plate capacitor?

Gauss' Law can be used to determine the electric field between the plates of a large area parallel plate capacitor by considering a Gaussian surface that encloses the plates. The electric flux through this surface can be calculated by using Gauss' Law, and by equating it to the total charge enclosed within the surface, the electric field can be determined.

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