Derive the expression for energy stored in a charged capacitor.
Hint: The relationship between capacitance, charge, and electric potential is key here. The energy stored in a capacitor is equal to the work done in accumulating charge within it. This involves the relationship between electric potential and work.
Formula Used:
The relationship between voltage ([latex]V[/latex]), charge ([latex]Q[/latex]), and capacitance ([latex]C[/latex]) is given by:
[latex]V = \frac{Q}{C}[/latex]
where:
[latex]V[/latex] is the electric potential,
[latex]C[/latex] is the capacitance, and
[latex]Q[/latex] is the charge on the capacitor.
The work done ([latex]dW[/latex]) to move a small charge [latex]dQ[/latex] into a capacitor at potential [latex]V[/latex] is:
[latex]dW = V \, dQ[/latex]
Solution:
Consider a capacitor in a circuit with voltage [latex]V[/latex]. The capacitance [latex]C[/latex] and stored charge \( Q \) relate as:
[latex]Q = C \cdot V[/latex]
Our aim is to find the energy stored in the capacitor.
According to electrostatic principles, the energy stored in a capacitor is equal to the work required to transfer the charge into the capacitor under potential [latex]V[/latex]. Thus:
[latex]dW = V \, dQ[/latex]
Substitute [latex]V = \frac{Q}{C}[/latex]:
[latex]dW = \frac{Q}{C} \, dQ[/latex]
To find the total work [latex]W[/latex] done in moving the charge [latex]Q[/latex] to the capacitor, integrate:
[latex]W = \int_0^Q \frac{Q}{C} \, dQ[/latex]
[latex]W = \frac{1}{C} \int_0^Q Q \, dQ[/latex]
Solving the integral:
[latex]W = \frac{1}{2} \frac{Q^2}{C}[/latex]
Thus, the energy [latex]U[/latex] stored in the capacitor is:
[latex]U = \frac{1}{2} \frac{Q^2}{C}[/latex]
Using [latex]Q = C \cdot V[/latex], we can also express this as:
[latex]U = \frac{1}{2} C V^2[/latex]
Additional Information:
A capacitor is a device that stores energy by separating equal and opposite charges with a certain distance between them. The standard capacitors shown in textbooks are often parallel-plate capacitors. Capacitors are measured in farads (F), where one farad stores one coulomb (C) of charge across one volt (V) of potential difference.
Note: For a parallel-plate capacitor, capacitance [latex]C[/latex] can also be calculated with the formula:
[latex]C = \frac{\varepsilon A}{d}[/latex]
where [latex]\varepsilon[/latex] is the permittivity of the medium between the plates, [latex]A[/latex] is the plate area, and [latex]d[/latex] is the distance between the plates. Substituting this expression for [latex]C[/latex] allows us to express energy in terms of the area and separation of the plates.