The image presents a problem about a parallel plate capacitor with the following details:

• The area of the plates $A = 0.04 \, \text{m}^2$
• The plates are separated by a distance $d = 1 \, \text{cm} = 0.01 \, \text{m}$
• The dielectric material between the plates has a relative permittivity $\varepsilon_r = 2$
• A potential difference of $V = 10 \, \text{V}$ is applied.

We are asked to find:

• (a) The capacitance $C$
• (b) The charge on the plates when a potential difference of 10 V is applied.
• (c) The energy stored in the capacitor.

Part (a): Capacitance Formula

The capacitance $C$ of a parallel plate capacitor with a dielectric material is given by the formula:

$C = \varepsilon_0 \varepsilon_r \frac{A}{d}$

Where:

• $\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}$ (vacuum permittivity),
• $\varepsilon_r = 2$ (relative permittivity),
• $A = 0.04 \, \text{m}^2$ (plate area),
• $d = 0.01 \, \text{m}$ (plate separation).

Substituting the values:

$C = (8.854 \times 10^{-12}) \times 2 \times \frac{0.04}{0.01}$

$C = 7.08 \times 10^{-12} \, \text{F} = 7.08 \, \text{pF}$

Part (b): Charge on the Plates

The charge $Q$ on the plates is related to the capacitance and voltage by:

$Q = C \cdot V$

Substituting the values:

$Q = 7.08 \times 10^{-12} \times 10 = 7.08 \times 10^{-11} \, \text{C} = 70.8 \, \text{pC}$

Part (c): Energy Stored

The energy $U$ stored in the capacitor is given by:

$U = \frac{1}{2} C V^2$

Substituting the values:

$U = \frac{1}{2} \times 7.08 \times 10^{-12} \times 10^2$

$U = 3.54 \times 10^{-10} \, \text{J}$

Thus, the results are:

• (a) Capacitance $C = 7.08 \, \text{pF}$
• (b) Charge $Q = 70.8 \, \text{pC}$
• (c) Energy stored $U = 3.54 \times 10^{-10} \, \text{J}$

Would you like further details or have any questions?

Related Questions:

1. How does the capacitance change if the separation between the plates is reduced?
2. What happens to the stored energy if the applied voltage is doubled?
3. How would the result differ if the dielectric constant $\varepsilon_r$ were 4 instead of 2?
4. What is the effect of increasing the plate area on capacitance and charge?
5. Can you derive the formula for energy stored in a capacitor?

Tip: The energy stored in a capacitor depends on both the voltage and the capacitance.