Electrostatics & Circuits MCQ & Objective Questions
Understanding "Electrostatics & Circuits" is crucial for students preparing for school and competitive exams in India. This topic not only forms a significant part of the syllabus but also features prominently in various objective questions and MCQs. By practicing these questions, students can enhance their grasp of concepts and improve their chances of scoring better in exams.
What You Will Practise Here
Fundamental concepts of electrostatics, including charge, electric field, and potential.
Key formulas related to Coulomb's law and electric field strength.
Understanding of capacitors, their types, and applications in circuits.
Basic circuit theory, including Ohm's law and Kirchhoff's laws.
Analysis of series and parallel circuits with practical examples.
Diagrams illustrating electric field lines and circuit schematics.
Problem-solving strategies for common electrostatics and circuit-related questions.
Exam Relevance
The topics of Electrostatics and Circuits are integral to the curriculum of CBSE, State Boards, NEET, and JEE. Students can expect questions that test their understanding of theoretical concepts as well as practical applications. Common question patterns include numerical problems, conceptual MCQs, and diagram-based questions that require a clear understanding of the subject matter.
Common Mistakes Students Make
Confusing the concepts of electric field and electric potential.
Misapplying Ohm's law in complex circuits.
Overlooking the significance of units in calculations.
Failing to interpret circuit diagrams accurately.
Neglecting to review the properties of capacitors and their behavior in circuits.
FAQs
Question: What are the key formulas I should remember for Electrostatics?Answer: Important formulas include Coulomb's law (F = k * |q1 * q2| / r²) and the formula for electric field (E = F/q).
Question: How can I improve my performance in circuit-related MCQs?Answer: Practice solving circuit problems regularly and familiarize yourself with different circuit configurations.
Question: Are there any specific topics I should focus on for competitive exams?Answer: Focus on understanding capacitors, circuit laws, and the relationship between voltage, current, and resistance.
Now is the time to boost your exam preparation! Dive into our practice MCQs on Electrostatics & Circuits and test your understanding to achieve your academic goals.
Q. What is the electric field strength at a distance of 2 m from a +5 µC point charge?
A.
1.125 N/C
B.
2.25 N/C
C.
0.5625 N/C
D.
0.75 N/C
Show solution
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:
B
— 2.25 N/C
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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
Show solution
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
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Q. What is the electric field strength at a distance of 2 m from a point charge of +5 µC?
A.
1.12 N/C
B.
0.56 N/C
C.
2.25 N/C
D.
0.75 N/C
Show solution
Solution
E = k * |q| / r^2 = (8.99 x 10^9 N m²/C²) * (5 x 10^-6 C) / (2 m)^2 = 1.12 N/C.
Correct Answer:
A
— 1.12 N/C
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Q. What is the electric field strength 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.
2000 N/C
Show solution
Solution
The electric field (E) is given by E = V/d. Here, E = 100V / 0.1m = 1000 N/C.
Correct Answer:
A
— 1000 N/C
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Q. What is the electric potential (V) across a capacitor (C) charged to a charge (Q)?
A.
V = Q/C
B.
V = C/Q
C.
V = Q*C
D.
V = C^2/Q
Show solution
Solution
The electric potential across a capacitor is given by the formula V = Q/C, where Q is the charge and C is the capacitance.
Correct Answer:
A
— V = Q/C
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Q. What is the electric potential (V) at a distance (r) from a point charge (Q)?
A.
V = k * Q / r
B.
V = k * Q * r
C.
V = Q / (4 * π * ε * r^2)
D.
V = Q / (4 * π * ε * r)
Show solution
Solution
The electric potential due to a point charge is given by V = k * Q / r, where k is Coulomb's constant.
Correct Answer:
A
— V = k * Q / r
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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
Show solution
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
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Q. What is the electric potential at a point 1 m away from a charge of +1 µC?
A.
9 kV
B.
1 kV
C.
0.9 kV
D.
0.1 kV
Show solution
Solution
V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) / 1 m = 8.99 kV.
Correct Answer:
A
— 9 kV
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Q. What is the electric potential at a point 1 m away from a charge of +3 µC?
A.
9 V
B.
27 V
C.
18 V
D.
36 V
Show solution
Solution
V = k * q / r = (8.99 x 10^9 N m²/C²) * (3 x 10^-6 C) / (1 m) = 26.97 V, approximately 27 V.
Correct Answer:
B
— 27 V
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Q. What is the electric potential at a point 3 m away from a charge of +1 µC?
A.
3000 V
B.
9000 V
C.
300 V
D.
900 V
Show solution
Solution
V = k * q / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) / 3 m = 3000 V.
Correct Answer:
C
— 300 V
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Q. What is the electric potential at a point 3 m away from a charge of +2 µC?
A.
0.6 V
B.
1.2 V
C.
2.4 V
D.
4.8 V
Show solution
Solution
V = k * q / r = (8.99 x 10^9 N m²/C²) * (2 x 10^-6 C) / 3 m = 5.99 V.
Correct Answer:
B
— 1.2 V
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Q. What is the electric potential energy of a charge of +2 µC placed in an electric field of 500 N/C at a distance of 0.1 m?
A.
0.1 mJ
B.
0.2 mJ
C.
0.3 mJ
D.
0.4 mJ
Show solution
Solution
Electric potential energy U = q * E * d = (2 x 10^-6 C) * (500 N/C) * (0.1 m) = 0.1 mJ.
Correct Answer:
B
— 0.2 mJ
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Q. What is the electric potential energy of a charge of +3 µC placed in an electric field of 2000 N/C at a distance of 0.5 m?
A.
3 J
B.
1.5 J
C.
0.3 J
D.
0.6 J
Show solution
Solution
Potential energy U = q * E * d = (3 x 10^-6 C) * (2000 N/C) * (0.5 m) = 3 J.
Correct Answer:
B
— 1.5 J
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Q. What is the electric potential energy of a system of two charges of +1 µC and -1 µC separated by 0.1 m?
A.
-0.09 J
B.
0.09 J
C.
0.18 J
D.
0.36 J
Show solution
Solution
Potential energy U = k * q1 * q2 / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) * (-1 x 10^-6 C) / 0.1 m = -0.09 J.
Correct Answer:
B
— 0.09 J
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Q. What is the electric potential energy of a system of two charges, +1 µC and -1 µC, separated by 0.1 m?
A.
-0.09 J
B.
0.09 J
C.
0.18 J
D.
0.36 J
Show solution
Solution
Potential energy U = k * q1 * q2 / r = (8.99 x 10^9 N m²/C²) * (1 x 10^-6 C) * (-1 x 10^-6 C) / 0.1 m = -0.09 J.
Correct Answer:
A
— -0.09 J
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Q. What is the electric potential energy stored in a capacitor of 2 µF charged to 12 V?
A.
0.144 mJ
B.
0.288 mJ
C.
0.576 mJ
D.
0.072 mJ
Show solution
Solution
U = 1/2 * C * V² = 1/2 * 2 x 10^-6 F * (12 V)² = 0.144 mJ.
Correct Answer:
B
— 0.288 mJ
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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
Show solution
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
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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
Show solution
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
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Q. What is the energy stored in a capacitor of 20 µF charged to 10 V?
A.
0.1 J
B.
0.2 J
C.
0.05 J
D.
0.15 J
Show solution
Solution
Energy U = 0.5 * C * V² = 0.5 * 20 x 10^-6 F * (10 V)² = 0.1 J.
Correct Answer:
B
— 0.2 J
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Q. What is the energy stored in a capacitor of 20 µF charged to 12 V?
A.
1.44 mJ
B.
0.12 mJ
C.
0.24 mJ
D.
0.48 mJ
Show solution
Solution
Energy stored, U = 0.5 * C * V^2 = 0.5 * (20 x 10^-6 F) * (12 V)^2 = 1.44 mJ.
Correct Answer:
A
— 1.44 mJ
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Q. What is the energy stored in a capacitor of 4 µF charged to a voltage of 12 V?
A.
0.288 mJ
B.
0.576 mJ
C.
0.144 mJ
D.
0.072 mJ
Show solution
Solution
Using the formula U = 1/2 * C * V^2 = 1/2 * 4 x 10^-6 F * (12 V)^2 = 0.288 mJ.
Correct Answer:
B
— 0.576 mJ
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Q. What is the energy stored in a capacitor of 5 µF charged to 12V?
A.
0.36 mJ
B.
0.72 mJ
C.
0.12 mJ
D.
0.24 mJ
Show solution
Solution
Energy U = 1/2 * C * V^2 = 1/2 * 5 x 10^-6 F * (12V)^2 = 0.36 mJ.
Correct Answer:
A
— 0.36 mJ
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Q. What is the energy stored in a capacitor of 5 µF charged to a voltage of 10 V?
A.
0.25 mJ
B.
0.5 mJ
C.
0.75 mJ
D.
1 mJ
Show solution
Solution
Energy (U) = 0.5 * C * V² = 0.5 * 5 x 10^-6 F * (10 V)² = 0.25 mJ.
Correct Answer:
A
— 0.25 mJ
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Q. What is the energy stored in a capacitor of 5 µF charged to a voltage of 12 V?
A.
0.36 mJ
B.
0.60 mJ
C.
0.72 mJ
D.
0.84 mJ
Show solution
Solution
Energy stored U = 0.5 * C * V² = 0.5 * 5 x 10^-6 F * (12 V)² = 0.36 mJ.
Correct Answer:
C
— 0.72 mJ
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Q. What is the energy stored in a capacitor of 5µF charged to a voltage of 10V?
A.
0.25 mJ
B.
0.5 mJ
C.
1 mJ
D.
2.5 mJ
Show solution
Solution
Energy (U) = 0.5 * C * V² = 0.5 * 5 x 10^-6 F * (10V)² = 0.25 mJ.
Correct Answer:
B
— 0.5 mJ
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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
Show solution
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
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Q. What is the equivalent capacitance of two capacitors (3 µF and 6 µF) in series?
A.
2 µF
B.
1.5 µF
C.
9 µF
D.
18 µF
Show solution
Solution
1/Ceq = 1/C1 + 1/C2 = 1/3 + 1/6 = 1/2. Therefore, Ceq = 2 µF.
Correct Answer:
B
— 1.5 µF
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Q. What is the equivalent capacitance of two capacitors of 4 µF and 6 µF connected in series?
A.
2.4 µF
B.
10 µF
C.
1.5 µF
D.
3.6 µF
Show solution
Solution
1/C_total = 1/C1 + 1/C2 = 1/4 µF + 1/6 µF = 1.2 µF, so C_total = 2.4 µF.
Correct Answer:
A
— 2.4 µF
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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
Show solution
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
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Q. What is the equivalent capacitance of two capacitors, 3 µF and 6 µF, connected in series?
A.
2 µF
B.
1 µF
C.
9 µF
D.
4 µF
Show solution
Solution
1/C_eq = 1/C1 + 1/C2 = 1/3 + 1/6 = 1/2. Therefore, C_eq = 2 µF.
Correct Answer:
A
— 2 µF
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