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 force between two point charges of +3 µC and -2 µC separated by a distance of 0.5 m?
A.
1.2 N
B.
0.6 N
C.
0.4 N
D.
0.8 N
Show solution
Solution
Using Coulomb's law, F = k * |q1 * q2| / r^2 = (8.99 x 10^9 N m²/C²) * |3 x 10^-6 C * -2 x 10^-6 C| / (0.5 m)² = 1.2 N.
Correct Answer:
A
— 1.2 N
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Q. What is the force on a charge of +1 µC placed in an electric field of 1000 N/C?
A.
0.001 N
B.
0.1 N
C.
1 N
D.
10 N
Show solution
Solution
F = q * E = (1 x 10^-6 C) * (1000 N/C) = 0.001 N.
Correct Answer:
B
— 0.1 N
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Q. What is the force on a charge of +1 µC placed in an electric field of 500 N/C?
A.
0.5 N
B.
1 N
C.
2 N
D.
0.2 N
Show solution
Solution
F = q * E = (1 x 10^-6 C) * (500 N/C) = 0.5 N.
Correct Answer:
A
— 0.5 N
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Q. What is the formula for calculating the current (I) in an RC circuit after a time (t) when a voltage (V) is applied?
A.
I = V/R
B.
I = V(1 - e^(-t/RC))
C.
I = V/R * e^(-t/RC)
D.
I = V * e^(-t/RC)
Show solution
Solution
The current in an RC circuit after a time t is given by I = V(1 - e^(-t/RC)), where V is the voltage, R is the resistance, and C is the capacitance.
Correct Answer:
B
— I = V(1 - e^(-t/RC))
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Q. What is the formula for the charge (Q) on a capacitor at time t during charging in an RC circuit?
A.
Q = C*V(1 - e^(-t/RC))
B.
Q = C*V*e^(-t/RC)
C.
Q = C*V*t
D.
Q = C*V*t^2
Show solution
Solution
The charge on a capacitor during charging in an RC circuit is given by Q = C*V(1 - e^(-t/RC)).
Correct Answer:
A
— Q = C*V(1 - e^(-t/RC))
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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
Show solution
Solution
The electric field between two parallel plates is given by E = V/d.
Correct Answer:
A
— E = V/d
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Q. What is the formula for the electric field (E) due to a point charge (Q) at a distance (r)?
A.
E = k * Q / r^2
B.
E = Q / (4 * π * ε * r^2)
C.
E = Q / r^2
D.
E = k * Q * r^2
Show solution
Solution
The electric field due to a point charge is given by E = k * Q / r^2, where k is Coulomb's constant.
Correct Answer:
A
— E = k * Q / r^2
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Q. What is the formula for the energy stored in a capacitor?
A.
U = 1/2 C V^2
B.
U = C V
C.
U = C V^2
D.
U = 1/2 Q V
Show solution
Solution
The energy (U) stored in a capacitor is given by U = 1/2 C V^2.
Correct Answer:
A
— U = 1/2 C V^2
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Q. What is the formula for the total capacitance (C_total) of capacitors in series?
A.
1/C_total = 1/C1 + 1/C2
B.
C_total = C1 + C2
C.
C_total = C1 * C2
D.
C_total = C1 - C2
Show solution
Solution
For capacitors in series, the total capacitance is given by 1/C_total = 1/C1 + 1/C2.
Correct Answer:
A
— 1/C_total = 1/C1 + 1/C2
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Q. What is the magnetic field strength at a distance of 0.1 m from a long straight wire carrying a current of 5 A?
A.
0.1 T
B.
0.01 T
C.
0.05 T
D.
0.02 T
Show solution
Solution
Using the formula B = (μ0 * I) / (2 * π * r), where μ0 = 4π x 10^-7 T m/A, B = (4π x 10^-7 * 5) / (2 * π * 0.1) = 0.01 T.
Correct Answer:
B
— 0.01 T
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Q. What is the potential difference (V) across a capacitor after it has been fully charged in an RC circuit?
A.
V = 0
B.
V = V0
C.
V = R * I
D.
V = C * I
Show solution
Solution
After a capacitor is fully charged in an RC circuit, the potential difference across it is equal to the source voltage V = V0.
Correct Answer:
B
— V = V0
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Q. What is the potential difference across a 10 µF capacitor charged to 5 V?
A.
0.05 J
B.
0.05 C
C.
0.05 V
D.
0.05 F
Show solution
Solution
The potential difference is simply the voltage, which is 5 V.
Correct Answer:
C
— 0.05 V
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Q. What is the potential difference across a 10Ω resistor carrying a current of 3A?
A.
10 V
B.
20 V
C.
30 V
D.
40 V
Show solution
Solution
Using Ohm's law, V = I * R = 3A * 10Ω = 30 V.
Correct Answer:
C
— 30 V
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Q. What is the potential difference across a 20 Ω resistor carrying a current of 2 A?
A.
40 V
B.
20 V
C.
10 V
D.
30 V
Show solution
Solution
Using Ohm's law, V = I * R = 2 A * 20 Ω = 40 V.
Correct Answer:
A
— 40 V
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Q. What is the potential difference across a 5 µF capacitor charged to 10 V?
A.
50 mJ
B.
0.05 J
C.
0.5 J
D.
5 J
Show solution
Solution
Energy stored in a capacitor, U = 0.5 * C * V^2 = 0.5 * (5 x 10^-6 F) * (10 V)^2 = 0.05 J.
Correct Answer:
B
— 0.05 J
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Q. What is the potential difference across a capacitor (C) charged to a charge (Q)?
A.
V = Q / C
B.
V = C / Q
C.
V = Q * C
D.
V = C * Q
Show solution
Solution
The potential difference across a capacitor is given by V = Q / C.
Correct Answer:
A
— V = Q / C
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Q. What is the potential difference across a capacitor if it has a capacitance of 5μF and stores a charge of 10μC?
Show solution
Solution
The potential difference V across a capacitor is given by V = Q/C. Therefore, V = 10μC / 5μF = 2V.
Correct Answer:
C
— 3V
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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
Show solution
Solution
Using C = Q/V, we find V = Q/C = 10 µC / 5 µF = 2 V.
Correct Answer:
A
— 2 V
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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
Show solution
Solution
Using the formula V = Q/C, V = 20 µC / 5 µF = 4 V.
Correct Answer:
A
— 4 V
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Q. What is the potential difference across a capacitor of 10 µF charged to 5 V?
A.
0.05 J
B.
0.1 J
C.
0.2 J
D.
0.15 J
Show solution
Solution
Energy stored in a capacitor U = 0.5 * C * V² = 0.5 * 10 x 10^-6 F * (5 V)² = 0.125 J.
Correct Answer:
B
— 0.1 J
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Q. What is the potential difference across a capacitor of 10 µF charged to 5V?
A.
0.05 V
B.
0.5 V
C.
5 V
D.
50 V
Show solution
Solution
The potential difference across a capacitor is equal to the voltage it is charged to, which is 5 V.
Correct Answer:
C
— 5 V
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Q. What is the potential difference across a capacitor of 10 µF when it stores a charge of 20 µC?
A.
2 V
B.
0.5 V
C.
1 V
D.
3 V
Show solution
Solution
Using the formula Q = C * V, V = Q / C = 20 x 10^-6 C / 10 x 10^-6 F = 2 V.
Correct Answer:
A
— 2 V
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Q. What is the potential difference across a capacitor of 2 microfarads charged to 10 volts?
A.
20 mC
B.
0.02 C
C.
0.02 mC
D.
0.2 C
Show solution
Solution
The charge (Q) on a capacitor is given by Q = C * V. Here, Q = 2 x 10^-6 F * 10 V = 20 x 10^-6 C = 20 mC.
Correct Answer:
A
— 20 mC
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Q. What is the potential difference across a capacitor of 2 µF charged to 10V?
A.
20 µC
B.
10 µC
C.
5 µC
D.
15 µC
Show solution
Solution
Q = C * V = 2 µF * 10V = 20 µC.
Correct Answer:
A
— 20 µC
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Q. What is the potential difference across a capacitor of 5 µF charged to 10 V?
A.
0.05 V
B.
0.5 V
C.
5 V
D.
50 V
Show solution
Solution
V = Q/C; Q = C * V = 5 µF * 10 V = 50 µC, thus the potential difference is 10 V.
Correct Answer:
C
— 5 V
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Q. What is the potential difference across a capacitor of 5 µF charged to 10V?
A.
5V
B.
10V
C.
15V
D.
20V
Show solution
Solution
The potential difference across a capacitor is equal to the voltage it is charged to, which is 10V.
Correct Answer:
B
— 10V
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Q. What is the potential difference across a capacitor of 5 µF charged to 12 V?
A.
0.06 C
B.
0.06 J
C.
0.06 V
D.
0.06 F
Show solution
Solution
The energy stored in a capacitor is U = 0.5 * C * V^2 = 0.5 * 5 x 10^-6 F * (12 V)^2 = 0.36 J.
Correct Answer:
B
— 0.06 J
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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
Show solution
Solution
V = Q / C = 0.01 C / 5 x 10^-6 F = 2000 V.
Correct Answer:
A
— 2 V
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Q. What is the potential difference across a capacitor of 8 µF charged with 16 µC?
Show solution
Solution
Voltage V = Q/C = 16 x 10^-6 C / 8 x 10^-6 F = 2V.
Correct Answer:
B
— 4V
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Q. What is the potential difference across a capacitor of 8 µF if it stores a charge of 16 µC?
Show solution
Solution
Using the formula Q = C * V, we have V = Q / C = 16 x 10^-6 C / 8 x 10^-6 F = 2V.
Correct Answer:
B
— 2V
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