This question aims to understand the concept of **capacitors,** how the electrical **charge** charges the capacitor, and how to calculate the **energy** stored in the capacitor.

In electrical **circuits,** the capacitor is commonly used as an **electrical** component, with storing an electrical **charge** as the main role. Charge of opposite **value** and same **magnitude** is present on adjacent **plates** in standard parallel plate **capacitors.** The electrical **potential** energy is stored in the capacitor.Â The **conductor** in the capacitor is initially uncharged and requires a **potential difference**Â **V**Â by connecting it to the battery. if at that time **q** is the charge on the plate, then **q = CV**. The product of the **potential** and **charge** is equal to the **work done.** Hence, **W = Vq**. The battery provides a little amount of **charge** at a stable **voltage** **V**, and the **energy stored** in the capacitor becomes:

\[ U = \dfrac{1}{2}CV^2\]

Applications of capacitors in microelectronics are **handheld** calculators, **audio** tools, **camera** flashes, **uninterruptible power** supplies, and **pulsed loads** such as magnetic coils and lasers.

## Expert Answer

**Part a:**

In this question, we are given:

The **capacitance** of the capacitor that is: $C \space=\space 8 \mu F$ and that is equal to: $\space 8 \times 10^{-6}$

The **energy** stored in the **capacitor** that is: $U_c \space=\space 40J$

And we are asked to find the **voltage** in the capacitor.

The formula that relates the **voltage** in the capacitor, the **capacitance** of the capacitor, and the **energy** stored in the capacitor are given as:

\[U_c=\dfrac{1}{2}V^2C\]

Rearranging the formula to make **Voltage** $V$ the subject because it is an unknown parameter that we are asked to find:

\[V=\sqrt{ \dfrac{2U_c}{C}}\]

Now plugging the values of $U_c$ and $C$ and **solving** for $V$:

\[ V= \sqrt{ \dfrac{2 \times 40}{8 \times 10^{-6}}} \]

By solving the **expression,** $V$ comes out to be:

\[ V=3.162 \space KV \]

**Part b:**

The stored **charge** $Q$ is the unknown parameter.

The formula that related the **energy** stored in the capacitor $U_c$, **Voltage** $V$ and the stored **charge** $Q$ is given as:

\[ U_c = \dfrac{1}{2}QV \]

Making $Q$ the subject:

\[ Q = \dfrac{2U_c}{v} \]

Plugging the **values** and **solving:**

\[ Q = \dfrac{2 \times 40}{3162} \]

By solving the **expression,** $Q$ comes out to be:

\[Q=0.0253 \space C\]

## Numerical Results

**Part a:** Voltage is applied to the $8.00 \mu F$ **capacitor** of a heart defibrillator that stores $40.0 J$ of **energy** is $3.16 \space KV$.

**Part b:**Â The **amount** of the stored **charge** is $0.0253C$.

**Ex**ample

A $12pF$ **capacitor** is connected to a $50V$ battery. Once the capacitor is fully **charged,** how much **electrostatic** energy is stored?

The formula given for finding the amount of **energy** stored in the capacitor is:

\[E \space = \space \dfrac{1}{2} CV^2\]

\[E \space = \space \dfrac{1}{2} (12 \times 10^{-12})(50)^2 \]

By **solving** the expression,Â **Energy** $E$ comes out to be:

\[E \space = 1.5 \times 10^{-8} J \]

Once the **capacitor** is fully charged, **electrostatic energy** stored is $ 1.5 \times 10^{-8} J$