Electrodes of area A are spaced distance d apart to form a parallel-plate capacitor. The electrodes are charged to ±q. What is the infinitesimal increase in electric potential energy dU if an infinitesimal amount of charge dq is moved from the negative electrode to the positive electrode?

Verified step by step guidance

Start by recalling the formula for the electric potential difference (V) between the plates of a parallel-plate capacitor: V = \( \frac{q}{C} \), where q is the charge on the capacitor and C is the capacitance.

The capacitance of a parallel-plate capacitor is given by \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 \) is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.

The infinitesimal increase in electric potential energy, dU, is related to the infinitesimal work done to move a small charge dq against the electric potential difference V. This is expressed as \( dU = V \cdot dq \).

Substitute the expression for V (\( V = \frac{q}{C} \)) into the equation for dU: \( dU = \frac{q}{C} \cdot dq \).

Finally, replace C with its expression (\( C = \frac{\varepsilon_0 A}{d} \)) to express dU in terms of q, dq, A, d, and \( \varepsilon_0 \): \( dU = \frac{q \cdot dq \cdot d}{\varepsilon_0 A} \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Potential Energy

Electric potential energy is the energy stored in an electric field due to the position of charged particles. In the context of capacitors, it is the work done to move a charge within the electric field created by the charged plates. The potential energy increases as charge is moved against the electric field, which is crucial for understanding how energy is stored in capacitors.

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is defined as C = q/V, where q is the charge stored and V is the voltage across the plates. The geometry of the capacitor, including the area of the plates and the distance between them, influences capacitance, which is essential for calculating changes in potential energy when charge is moved.

Electric Field

The electric field (E) between the plates of a capacitor is a vector field that represents the force per unit charge experienced by a positive test charge placed in the field. It is uniform in a parallel-plate capacitor and is given by E = V/d, where V is the voltage and d is the distance between the plates. Understanding the electric field is vital for determining the work done when moving charge and thus the change in electric potential energy.

Recommended video:

Guided course

03:16

Intro to Electric Fields

Related Practice

Textbook Question

Two 10-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. After a long time, the capacitor is disconnected from the battery but is not discharged. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after the original electrodes (not the modified electrodes of part b) are expanded until they are 20 cm in diameter?

Textbook Question

Two spherical drops of mercury each have a charge of 0.10 nC and a potential of 300 V at the surface. The two drops merge to form a single drop. What is the potential at the surface of the new drop?

views

Textbook Question

A 2.0-cm-diameter copper ring has 5.0×10⁹ excess electrons. A proton is released from rest on the axis of the ring, 5.0 cm from its center. What is the proton's speed as it passes through the center of the ring?

Textbook Question

The potential 1.0 cm from the surface of a metal sphere is 8000 V. The potential 3.0 cm from the surface is 4000 V. What is the radius of the sphere?

views

Textbook Question

In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons, thus having mass m=4 u and charge q=2e. Suppose a uranium nucleus with 92 protons decays into thorium, with 90 protons, and an alpha particle. The alpha particle is initially at rest at the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest.

views

Textbook Question

A Van de Graaff generator is a device for generating a large electric potential by building up charge on a hollow metal sphere. A typical classroom-demonstration model has a diameter of 30 cm. What is the electric field strength just outside the surface of the sphere when it is charged to 500,000 V?