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?

Verified step by step guidance

1. Step 1: Understand the problem. The problem involves a parallel-plate capacitor with initial dimensions and voltage. After being disconnected from the battery, the plates are expanded to a new diameter. We need to calculate the charge on each electrode, the electric field strength, and the potential difference after the expansion.
2. Step 2: Recall the formula for capacitance of a parallel-plate capacitor: $C=\frac{\epsilon }{A}/d$, where $\epsilon$ is the permittivity of free space, $A$ is the area of the plates, and $d$ is the separation between the plates. Calculate the initial capacitance using the given diameter and separation.
3. Step 3: Use the relationship between charge, capacitance, and voltage: $Q=CV$. Since the capacitor is disconnected from the battery, the charge $Q$ remains constant even after the plates are expanded. Calculate the initial charge using the initial capacitance and voltage.
4. Step 4: After the plates are expanded, the area $A$ changes, which affects the capacitance. Recalculate the new capacitance using the new diameter and the same separation distance. Use the formula for capacitance again.
5. Step 5: Determine the new potential difference using the relationship $V=\frac{Q}{C}$. Also, calculate the electric field strength inside the capacitor using the formula $E=\frac{V}{d}$, where $d$ is the separation distance.

Key Concepts

Capacitance

Capacitance is the ability of a system to store electric charge per unit voltage. For parallel-plate capacitors, it is given by the formula C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the area of one plate, and d is the separation between the plates. The larger the area and the smaller the distance between the plates, the greater the capacitance.

Electric Field

The electric field (E) between the plates of a capacitor is defined as the force per unit charge experienced by a positive test charge placed in the field. For a parallel-plate capacitor, the electric field is uniform and can be calculated using E = V/d, where V is the potential difference and d is the separation between the plates. This field is crucial for understanding how charges interact within the capacitor.

Charge Distribution

When a capacitor is connected to a voltage source, charge accumulates on its plates, creating an electric field. The amount of charge (Q) on each plate is directly proportional to the capacitance and the voltage applied, expressed as Q = C × V. After disconnecting from the battery, the charge remains constant, and any changes in the physical dimensions of the capacitor, such as increasing the plate diameter, will affect the capacitance and electric field strength.