Question

Deduce the equation of potential energy of a charged condenser and show that the energy density in the electric field between the plates of charged condenser is \(\frac{1}{2} \epsilon_0 E^2\).

Hint:

The three forms of capacitor energy (\(U = \frac{1}{2}CV^2\), \(U = \frac{Q^2}{2C}\), \(U = \frac{1}{2}QV\)) are all equivalent. Choose the most convenient one based on the quantities given in a problem. The energy density formula is general and applies to any electric field, not just the one in a capacitor.

Answer 1

Part 1: Deduction of Potential Energy of a Capacitor

Step 1: Understanding the Concept:
The potential energy stored in a capacitor is equal to the total work done in charging the capacitor. This work is done by an external agent (like a battery) to transfer charge from one plate to another against the electrostatic forces.

Step 2: Derivation:
Let's consider a capacitor of capacitance \(C\). At an intermediate stage during charging, let \(q\) be the charge on the plates and \(V\) be the potential difference between them. By definition, \(V = q/C\).
The work done (\(dW\)) to transfer an additional infinitesimal charge \(dq\) from the negative plate to the positive plate is given by: \[ dW = V \, dq \] Substituting \(V = q/C\): \[ dW = \frac{q}{C} \, dq \] To find the total work done (\(W\)) in charging the capacitor from an initial charge of 0 to a final charge of \(Q\), we integrate \(dW\): \[ W = \int dW = \int_{0}^{Q} \frac{q}{C} \, dq \] Since \(C\) is a constant: \[ W = \frac{1}{C} \int_{0}^{Q} q \, dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_{0}^{Q} = \frac{1}{C} \left( \frac{Q^2}{2} - 0 \right) = \frac{Q^2}{2C} \] This work done is stored as electrostatic potential energy (\(U\)) in the capacitor. \[ U = \frac{Q^2}{2C} \] Using the relation \(Q = CV\), we can express the energy in other forms: \begin{itemize} \item \(U = \frac{(CV)^2}{2C} = \frac{1}{2}CV^2\) \item \(U = \frac{Q^2}{2(Q/V)} = \frac{1}{2}QV\) \end{itemize} Thus, the potential energy is \(U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV\).
Part 2: Derivation of Energy Density

Step 1: Understanding the Concept:
Energy density (\(u_E\)) is the electrostatic potential energy stored per unit volume in the electric field.

Step 2: Derivation:
Consider a parallel plate capacitor with plate area \(A\) and separation distance \(d\). The volume of the space between the plates is \(Volume = A \times d\).
The capacitance of this capacitor is given by: \[ C = \frac{\epsilon_0 A}{d} \] The electric field \(E\) between the plates is uniform and is related to the potential difference \(V\) by: \[ E = \frac{V}{d} \implies V = Ed \] Now, let's use the potential energy formula \(U = \frac{1}{2}CV^2\) and substitute the expressions for \(C\) and \(V\): \[ U = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) (Ed)^2 = \frac{1}{2} \frac{\epsilon_0 A}{d} E^2 d^2 \] \[ U = \frac{1}{2} \epsilon_0 E^2 (A \cdot d) \] Energy density (\(u_E\)) is the energy per unit volume: \[ u_E = \frac{\text{Energy (U)}}{\text{Volume}} = \frac{U}{A \cdot d} \] \[ u_E = \frac{\frac{1}{2} \epsilon_0 E^2 (A \cdot d)}{A \cdot d} \] \[ u_E = \frac{1}{2} \epsilon_0 E^2 \] This shows that the energy density in the electric field between the plates of a charged capacitor is \(\frac{1}{2} \epsilon_0 E^2\).