Electric Potential and Capacitance for Competitive Exams

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Electric Potential and Capacitance - Capacitance and Dielectrics

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Electric Potential and Capacitance - Capacitance and Dielectrics MCQ & Objective Questions

The topic of "Electric Potential and Capacitance - Capacitance and Dielectrics" is crucial for students preparing for school and competitive exams. Understanding these concepts not only enhances your grasp of physics but also significantly boosts your performance in exams. Practicing MCQs and objective questions related to this topic helps in reinforcing your knowledge and identifying important questions that frequently appear in exams.

What You Will Practise Here

Definition and significance of electric potential and capacitance
Key formulas related to capacitance and dielectrics
Understanding the concept of dielectric materials and their properties
Capacitance in series and parallel combinations
Energy stored in a capacitor and its derivation
Applications of capacitors in real-life scenarios
Diagrams illustrating electric field lines and equipotential surfaces

Exam Relevance

This topic is frequently featured in CBSE, State Boards, NEET, and JEE exams. Students can expect questions that test their understanding of the basic principles of capacitance, calculations involving capacitors, and the application of dielectrics in various scenarios. Common question patterns include numerical problems, conceptual questions, and application-based queries, making it essential to master this area for exam success.

Common Mistakes Students Make

Confusing the definitions of electric potential and electric field
Miscalculating capacitance in series and parallel circuits
Overlooking the effects of dielectrics on capacitance
Neglecting to apply the correct formulas in numerical problems
Failing to interpret diagrams accurately, leading to incorrect answers

FAQs

Question: What is the formula for capacitance?
Answer: The capacitance (C) is defined as C = Q/V, where Q is the charge stored and V is the voltage across the capacitor.

Question: How does a dielectric affect capacitance?
Answer: A dielectric material increases the capacitance of a capacitor by reducing the electric field within it, allowing more charge to be stored for the same voltage.

Now is the time to enhance your understanding of "Electric Potential and Capacitance - Capacitance and Dielectrics." Dive into our practice MCQs and test your knowledge to ensure you are well-prepared for your exams. Remember, consistent practice is the key to success!

Q. A capacitor is charged to a voltage of 15 V and then disconnected from the battery. If the capacitance is 3 µF, what is the charge on the capacitor?

A. 45 µC
B. 30 µC
C. 15 µC
D. 60 µC

Solution

Charge (Q) is given by Q = CV. Here, Q = 3 µF * 15 V = 45 µC.

Correct Answer: A — 45 µC

Q. A capacitor is charged to a voltage of 15 V and then disconnected from the power supply. What happens to the charge on the capacitor if the voltage is doubled?

A. Charge doubles
B. Charge halves
C. Charge remains the same
D. Charge becomes zero

Solution

If the voltage is doubled, the charge on the capacitor also doubles, since Q = C * V.

Correct Answer: A — Charge doubles

Q. If a capacitor is charged to a voltage of 20 V and then connected in parallel with another uncharged capacitor of the same capacitance, what is the new voltage across both capacitors?

A. 10 V
B. 20 V
C. 30 V
D. 40 V

Solution

When two identical capacitors are connected in parallel, the charge is shared equally. The new voltage is halved: 20 V / 2 = 10 V.

Correct Answer: A — 10 V

Q. If a capacitor is connected to an AC source, what happens to the current through the capacitor?

A. It is constant
B. It leads the voltage
C. It lags the voltage
D. It is zero

Solution

In an AC circuit, the current through a capacitor leads the voltage by 90 degrees.

Correct Answer: B — It leads the voltage

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

A. It doubles
B. It quadruples
C. It halves
D. It remains the same

Solution

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

Correct Answer: B — It quadruples

Q. If the capacitance of a capacitor is 3 µF and it is charged to 6 V, what is the charge on the capacitor?

A. 18 µC
B. 12 µC
C. 6 µC
D. 9 µC

Solution

Charge (Q) is given by Q = C * V. Here, Q = 3 µF * 6 V = 18 µC.

Correct Answer: A — 18 µC

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

A. 10 Ω
B. 20 Ω
C. 30 Ω
D. 5 Ω

Solution

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

Correct Answer: C — 30 Ω

Q. In a series circuit with a 12 V battery and two resistors of 4 Ω and 6 Ω, what is the voltage across the 4 Ω resistor?

A. 4.8 V
B. 6 V
C. 8 V
D. 3.6 V

Solution

Total resistance R = 4 Ω + 6 Ω = 10 Ω. Current I = V/R = 12 V / 10 Ω = 1.2 A. Voltage across 4 Ω = I * R = 1.2 A * 4 Ω = 4.8 V.

Correct Answer: A — 4.8 V

Q. In a series circuit with two capacitors of capacitance 3 µF and 6 µF, what is the total capacitance?

A. 2 µF
B. 1 µF
C. 9 µF
D. 4 µF

Solution

Total capacitance in series is given by 1/C_total = 1/C1 + 1/C2. Thus, 1/C_total = 1/3 + 1/6 = 1/2, so C_total = 2 µF.

Correct Answer: A — 2 µF

Q. What happens to the capacitance of a capacitor if the plate area is doubled while keeping the distance constant?

A. It doubles
B. It halves
C. It quadruples
D. It remains the same

Solution

Capacitance is directly proportional to the plate area. Doubling the area doubles the capacitance.

Correct Answer: A — It doubles

Q. What is the capacitance of a capacitor with a charge of 5 µC and a potential difference of 10 V?

A. 0.5 µF
B. 1 µF
C. 2 µF
D. 0.2 µF

Solution

Capacitance (C) is given by C = Q/V. Here, C = 5 µC / 10 V = 0.5 µF.

Correct Answer: A — 0.5 µF

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

A. Increases capacitance
B. Decreases capacitance
C. No effect on capacitance
D. Increases resistance

Solution

Adding a dielectric increases the capacitance of the capacitor.

Correct Answer: A — Increases capacitance

Q. What is the effect of connecting a capacitor in parallel with a resistor in a DC circuit?

A. Increases current
B. Decreases current
C. Stores energy
D. No effect

Solution

A capacitor stores energy when connected in parallel with a resistor, affecting the circuit's behavior over time.

Correct Answer: C — Stores energy

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

A. Capacitance increases
B. Capacitance decreases
C. Capacitance remains the same
D. Capacitance becomes zero

Solution

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

Correct Answer: B — Capacitance decreases

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

A. Capacitance increases
B. Capacitance decreases
C. Capacitance remains the same
D. Capacitance becomes zero

Solution

Capacitance decreases as the distance between the plates increases.

Correct Answer: B — Capacitance decreases

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

A. 0.144 mJ
B. 0.024 mJ
C. 0.288 mJ
D. 0.072 mJ

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.

Correct Answer: A — 0.144 mJ

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

A. 0.144 mJ
B. 0.12 mJ
C. 0.24 mJ
D. 0.06 mJ

Solution

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

Correct Answer: A — 0.144 mJ

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

A. 0.25 mJ
B. 0.5 mJ
C. 0.75 mJ
D. 1 mJ

Solution

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

Correct Answer: B — 0.5 mJ

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

A. 1.2 µF
B. 5 µF
C. 0.6 µF
D. 6 µF

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.

Correct Answer: A — 1.2 µF

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

A. E = V/d
B. E = d/V
C. E = V*d
D. E = d^2/V

Solution

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

Correct Answer: A — E = V/d

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

A. 2 V
B. 5 V
C. 10 V
D. 20 V

Solution

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

Correct Answer: A — 2 V

Q. What is the potential difference across a capacitor if it stores 20 µC of charge and has a capacitance of 5 µF?

A. 4 V
B. 5 V
C. 2 V
D. 10 V

Solution

Using the formula V = Q/C, V = 20 µC / 5 µF = 4 V.

Correct Answer: A — 4 V

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