Coulombs Law and Electric Field for Competitive Exams

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Coulombs Law and Electric Field

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Coulombs Law and Electric Field MCQ & Objective Questions

Coulombs Law and Electric Field are fundamental concepts in physics that play a crucial role in various school and competitive exams. Understanding these topics not only enhances your conceptual clarity but also boosts your confidence in tackling objective questions. Practicing MCQs related to Coulombs Law and Electric Field is essential for effective exam preparation, as it helps you identify important questions and reinforces your learning through practice questions.

What You Will Practise Here

Understanding Coulombs Law and its mathematical expression.
Calculating electric fields due to point charges and charge distributions.
Exploring the concept of electric field lines and their properties.
Applying superposition principle in electric fields.
Analyzing the relationship between electric field and force on a charge.
Solving numerical problems involving Coulombs Law and electric field strength.
Identifying and interpreting diagrams related to electric fields.

Exam Relevance

The topics of Coulombs Law and Electric Field are frequently included in the syllabus of CBSE, State Boards, NEET, and JEE. Students can expect questions that test both theoretical understanding and practical application. Common question patterns include numerical problems, conceptual MCQs, and diagram-based questions that require students to analyze electric field configurations. Mastery of these concepts is vital for scoring well in physics exams.

Common Mistakes Students Make

Confusing the direction of electric field lines and force on a charge.
Misapplying Coulombs Law in multi-charge systems.
Overlooking the significance of charge signs when calculating electric fields.
Neglecting to consider the superposition principle in complex problems.
Failing to interpret electric field diagrams accurately.

FAQs

Question: What is Coulombs Law?
Answer: Coulombs Law describes the force between two point charges, stating that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Question: How do you calculate the electric field due to a point charge?
Answer: The electric field (E) due to a point charge (Q) is calculated using the formula E = k * |Q| / r², where k is Coulomb's constant and r is the distance from the charge.

Now is the time to enhance your understanding of these crucial concepts! Dive into solving practice MCQs and test your knowledge on Coulombs Law and Electric Field. Every question you tackle brings you one step closer to exam success!

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

A. 0.125 mJ
B. 0.25 mJ
C. 0.5 mJ
D. 1 mJ

Solution

Energy stored, U = 0.5 * C * V² = 0.5 * 10 x 10^-6 F * (5 V)² = 0.125 mJ.

Correct Answer: B — 0.25 mJ

Q. If the electric potential at a point is 100 V and the charge at that point is 2 µC, what is the electric potential energy?

A. 0.2 J
B. 0.1 J
C. 0.05 J
D. 0.4 J

Solution

Potential energy U = V * q = 100 V * 2 x 10^-6 C = 0.0002 J = 0.2 J.

Correct Answer: A — 0.2 J

Q. If the potential difference across a capacitor is 12 V and its capacitance is 3 µF, what is the charge stored in the capacitor?

A. 36 µC
B. 24 µC
C. 12 µC
D. 18 µC

Solution

Q = C * V = 3 x 10^-6 F * 12 V = 36 µC.

Correct Answer: A — 36 µC

Q. If the potential difference across a capacitor is doubled, what happens to the stored energy?

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

Solution

Energy stored in a capacitor is given by U = 1/2 C V^2. If V is doubled, U becomes 1/2 C (2V)^2 = 2CV^2, which is quadrupled.

Correct Answer: B — It quadruples

Q. If two capacitors of 4 µF and 6 µF are connected in series, what is the total capacitance?

A. 2.4 µF
B. 10 µF
C. 1.5 µF
D. 24 µF

Solution

1/C_total = 1/C1 + 1/C2 = 1/4 + 1/6 = 5/12, thus C_total = 12/5 = 2.4 µF.

Correct Answer: A — 2.4 µF

Q. If two charges of +3 µC and -3 µC are placed 0.2 m apart, what is the magnitude of the force between them?

A. 67.5 N
B. 45 N
C. 30 N
D. 15 N

Solution

Using Coulomb's law, F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * |3 x 10^-6 C * -3 x 10^-6 C| / (0.2 m)^2 = 45 N.

Correct Answer: B — 45 N

Q. In a parallel circuit with a 12 V battery and two resistors of 4 Ω and 6 Ω, what is the total current supplied by the battery?

A. 2 A
B. 3 A
C. 4 A
D. 1 A

Solution

Total resistance, R_eq = 1/(1/4 + 1/6) = 2.4 Ω. Total current, I = V/R_eq = 12 V / 2.4 Ω = 5 A.

Correct Answer: B — 3 A

Q. In a parallel circuit with a total voltage of 12 V, what is the voltage across each resistor?

A. 6 V
B. 12 V
C. 24 V
D. 0 V

Solution

In a parallel circuit, the voltage across each resistor is equal to the total voltage, so it is 12 V.

Correct Answer: B — 12 V

Q. Two charges, +4 µC and -4 µC, are placed 1 m apart. What is the net electric field at the midpoint between the charges?

A. 0 N/C
B. 8.99 N/C
C. 4.49 N/C
D. 17.98 N/C

Solution

The electric fields due to both charges at the midpoint are equal in magnitude and opposite in direction, resulting in a net electric field of 0 N/C.

Correct Answer: A — 0 N/C

Q. What is the capacitance of a capacitor that stores 0.01 C of charge at a potential difference of 5 V?

A. 0.002 F
B. 0.01 F
C. 0.005 F
D. 0.1 F

Solution

C = Q / V = 0.01 C / 5 V = 0.002 F.

Correct Answer: A — 0.002 F

Q. What is the capacitance of a capacitor that stores 0.01 J of energy at a voltage of 10 V?

A. 0.2 F
B. 0.1 F
C. 0.05 F
D. 0.01 F

Solution

Using U = 1/2 C V^2, rearranging gives C = 2U/V^2 = 2(0.01 J)/(10 V)^2 = 0.1 F.

Correct Answer: B — 0.1 F

Q. What is the capacitance of a capacitor that stores 20 µC of charge at a potential difference of 5 V?

A. 2 µF
B. 4 µF
C. 5 µF
D. 10 µF

Solution

C = Q/V = 20 x 10^-6 C / 5 V = 4 µF.

Correct Answer: B — 4 µF

Q. What is the electric field between two parallel plates separated by 0.1 m with a potential difference of 100 V?

A. 1000 N/C
B. 500 N/C
C. 100 N/C
D. 10 N/C

Solution

E = V/d = 100 V / 0.1 m = 1000 N/C.

Correct Answer: A — 1000 N/C

Q. What is the electric field due to a point charge of -4 µC at a distance of 0.25 m?

A. -5760 N/C
B. -1440 N/C
C. 5760 N/C
D. 1440 N/C

Solution

E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (4 x 10^-6 C) / (0.25 m)^2 = -5760 N/C.

Correct Answer: A — -5760 N/C

Q. What is the electric field strength at a distance of 2 m from a charge of +5 µC?

A. 1.125 N/C
B. 2.25 N/C
C. 3.75 N/C
D. 4.5 N/C

Solution

E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * |5 x 10^-6 C| / (2 m)^2 = 1.125 N/C.

Correct Answer: A — 1.125 N/C

Q. What is the electric potential at a point 1 m away from a +1 C charge?

A. 9 N/C
B. 1 V
C. 8.99 V
D. 0 V

Solution

Electric potential V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 C) / (1 m) = 8.99 V.

Correct Answer: C — 8.99 V

Q. What is the equivalent resistance of three 4 Ω resistors in series?

A. 12 Ω
B. 8 Ω
C. 16 Ω
D. 4 Ω

Solution

R_eq = R1 + R2 + R3 = 4 Ω + 4 Ω + 4 Ω = 12 Ω.

Correct Answer: A — 12 Ω

Q. What is the equivalent resistance of three resistors of 4 Ω, 6 Ω, and 12 Ω in series?

A. 22 Ω
B. 18 Ω
C. 16 Ω
D. 12 Ω

Solution

In series, R_eq = R1 + R2 + R3 = 4 Ω + 6 Ω + 12 Ω = 22 Ω.

Correct Answer: A — 22 Ω

Q. What is the potential difference across a capacitor of 5 µF charged with 0.01 C?

A. 2 V
B. 0.5 V
C. 5 V
D. 10 V

Solution

V = Q / C = 0.01 C / 5 x 10^-6 F = 2000 V.

Correct Answer: A — 2 V

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