Derive the formula for the energy stored in a capacitor, showing at least two forms.

When an infinitesimal charge $dq$ is transferred across a potential difference $V$, the work done $dW$ is given by:

$$dW = V dq$$

The capacitance $C$ of a capacitor is defined as the ratio of the charge $q$ stored on its plates to the potential difference $V$ across them:

$$C = \frac{q}{V}$$

From this, we can express $V$ as:

$$V = \frac{q}{C}$$

Substitute the expression for $V$ into the equation for $dW$:

$$dW = \frac{q}{C} dq$$

To find the total energy $W$ stored when the capacitor is charged from an initial charge of 0 to a final charge of $q$, we integrate $dW$:

$$W = \int_{0}^{q} \frac{q'}{C} dq'$$

Here, $q'$ is used as the integration variable to distinguish it from the upper limit $q$.

$$W = \frac{1}{C} \int_{0}^{q} q' dq'$$ $$W = \frac{1}{C} \left[ \frac{(q')^2}{2} \right]_{0}^{q}$$ $$W = \frac{1}{C} \left( \frac{q^2}{2} - \frac{0^2}{2} \right)$$ $$W = \frac{q^2}{2C}$$

This is the first form of the energy stored.

We know that $q = CV$. Substitute this into the expression $W = \frac{q^2}{2C}$:

$$W = \frac{(CV)^2}{2C} = \frac{C^2V^2}{2C} = \frac{1}{2}CV^2$$

This is the second form of the energy stored.

Now, substitute $C = \frac{q}{V}$ into the expression $W = \frac{1}{2}CV^2$:

$$W = \frac{1}{2} \left(\frac{q}{V}\right) V^2 = \frac{1}{2} qV$$

This is the third form of the energy stored.

Thus, the electrical energy stored in a capacitor can be expressed in three equivalent forms:

$$W = \frac{1}{2}CV^2 = \frac{q^2}{2C} = \frac{1}{2}qV$$