Question

A parallel-plate capacitor consists of two flat metal plates of area \(A\) separated by a small distance \(d\). The plates are given equal and opposite net charges \(\pm q\) (a) Sketch the field lines and use your sketch to explain why almost all of the charge is on the inner surfaces of the plates. (b) Use Gauss's law to show that the electric field between the plates and away from the edges is $E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} \cdot(\mathrm{c})$ Does this agree with or contra- dict the result of Problem \(70 ?\) Explain. (d) Use the principle of superposition and the result of Problem 69 to arrive at this same answer. [Hint: The inner surfaces of the two plates are thin, flat sheets of charge.]

Short answer

Answer: In a parallel-plate capacitor, the charge is primarily on the inner surfaces of the plates because the electric field lines connecting positive and negative charges are concentrated in the region between the plates, making the field stronger and more uniform in that region. On the outer surfaces of the plates, there is no opposing charge to attract the electric field lines, so the field lines spread out, leaving only a small portion of the charge on the outer surfaces.

Step-by-step solution

Sketch the electric field lines and discuss charge distribution

To draw the electric field lines, we need to remember that they originate from positive charges and terminate on negative charges. Between the two plates, the electric field lines will be straight and parallel, connecting positive and negative charges. They don't cross each other. On the outer surfaces of the plates, there is no other opposing charge to attract the electric field lines. Therefore, the field lines spread out, leaving only a small portion of the charge on the outer surfaces.

Use Gauss's law to find the electric field between the plates

We can apply Gauss's law to a cylindrical Gaussian surface, where one flat face is inside the capacitor and the other is outside. The Gaussian surface encloses the inner surface of one of the capacitor plates, thus including its charge. According to Gauss's law, \(\oint \vec{E} \cdot d\vec{A} = Q_{enc}/\epsilon_0\), where \(Q_{enc}\) is the enclosed charge. Between the plates, the electric field is uniform and perpendicular to the plate surfaces. Thus, the flux through the top and bottom faces of the Gaussian surface adds up, while the flux through the side surface becomes zero, as the electric field is parallel to the side surface. Then, the total electric flux is \(EA = Q_{enc}/\epsilon_0\). As we know, \(Q_{enc} = q\), and after rearranging the equation, the electric field between the plates is \(E = q/(\epsilon_0 A)\).

Express the electric field in terms of charge density

The charge density \(\sigma\) is defined as the charge per unit area: \(\sigma = q/A\). We can substitute this into the expression for the electric field, obtaining \(E = \sigma / \epsilon_0\).

Compare the results with previous problem results

The result found in this exercise (\(E=\sigma/\epsilon_0\)) agrees with the result of Problem 70. This shows that Gauss's law is consistent with previous findings for the electric field.

Utilize the principle of superposition and the result from Problem 69

By applying the principle of superposition, we can consider the electric field produced by each capacitor plate separately. The electric field from a uniformly charged flat sheet is given by \(E = \frac{1}{2} \frac{\sigma}{\epsilon_0}\) (from Problem 69). In the region between the plates, the electric fields due to the positive and negative charged sheets add up, yielding a total electric field of \(E = \frac{1}{2} \frac{\sigma}{\epsilon_0} + \frac{1}{2} \frac{\sigma}{\epsilon_0} = \frac{\sigma}{\epsilon_0}\). This result confirms the answer obtained through Gauss's law.