Electric Potential and Capacitance - Capacitance and Dielectrics

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Q1. What is the electric potential energy stored in a capacitor of capacitance 2 µF charged to 12 V?

Solution:

Electric potential energy (U) is given by U = 1/2 CV^2. Here, U = 1/2 * 2 µF * (12 V)^2 = 0.144 mJ.

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Q2. If a dielectric material with a dielectric constant of 4 is inserted into a capacitor, how does the capacitance change?

Solution:

The capacitance increases by a factor equal to the dielectric constant. Thus, it quadruples.

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Q3. What is the equivalent capacitance of two capacitors, 2 µF and 3 µF, connected in series?

Solution:

For capacitors in series, 1/C_eq = 1/C1 + 1/C2. Thus, 1/C_eq = 1/2 + 1/3 = 5/6. Therefore, C_eq = 6/5 = 1.2 µF.

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Q4. What is the energy stored in a capacitor of capacitance 5 µF charged to 10 V?

Solution:

Energy (U) is given by U = 1/2 CV^2. Here, U = 1/2 * 5 µF * (10 V)^2 = 0.5 mJ.

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Q5. What is the potential difference across a capacitor if it stores 10 µC of charge and has a capacitance of 5 µF?

Solution:

Using C = Q/V, we find V = Q/C = 10 µC / 5 µF = 2 V.

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Q6. What is the effect of increasing the distance between the plates of a parallel plate capacitor on its capacitance?

Solution:

Capacitance (C) is inversely proportional to the distance (d) between the plates. Increasing d decreases C.

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Q7. In a series circuit with a 10 Ω and a 20 Ω resistor, what is the total resistance?

Solution:

Total resistance in series is R_total = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.

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Q8. What is the effect of adding a dielectric to a capacitor?

Solution:

Adding a dielectric increases the capacitance of the capacitor.

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Q9. What is the effect of increasing the distance between the plates of a parallel plate capacitor?

Solution:

Capacitance decreases as the distance between the plates increases.

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Q10. What is the formula for the electric field (E) between two parallel plates separated by a distance (d) with a potential difference (V)?

Solution:

The electric field between two parallel plates is given by E = V/d.