Unit 11: Electrostatics — Online MCQ Test
PHYSICS · JEE MAINS · JEE
Practice Unit 11: Electrostatics with a free chapter-wise online MCQ test.
This chapter covers: Electric charges: conservation of charge Coulomb's law forces between twopoint charges forces between multiple charges superposition principle andcontinuous charge distribution. El....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Unit 11: Electrostatics — Important Questions & Answers
An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity 10^6 m/s. If the magnitude of the electric field between the plates is 9.1 V/cm, then the vertical component of velocity of electron is (mass of electron = 9.1 x 10^-31 kg and charge of electron = 1.6 x 10^-19 C)
- A. 0
- B. 1.6 x 10^6 m/s
- C. 1.6 x 10^4 m/s
- D. 1.6 x 10^5 m/s
Answer: C. 1.6 x 10^4 m/s
The time spent in the field t = length / velocity_horizontal = 0.1 / 10^6 = 10^-7 s. The acceleration a = qE/m. Given E = 9.1 V/cm = 910 V/m. a = (1.6 * 10^-19 * 910) / (9.1 * 10^-31) = 1.6 * 10^13 m/s^2. Vertical velocity v_y = a * t = 1.6 * 10^13 * 10^-7 = 1.6 * 10^6 m/s.
The time spent in the field t = length / velocity_horizontal = 0.1 / 10^6 = 10^-7 s. The acceleration a = qE/m. Given E = 9.1 V/cm = 910 V/m. a = (1.6 * 10^-19 * 910) / (9.1 * 10^-31) = 1.6 * 10^13 m/s^2. Vertical velocity v_y = a * t = 1.6 * 10^13 * 10^-7 = 1.6 * 10^6 m/s.
A parallel-plate capacitor of capacitance 40 μF is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are
- A. 4 mC and 0.2 J
- B. 8 mC and 2.0 J
- C. 2 mC and 0.4 J
- D. 2 mC and 0.2 J
Answer: A. 4 mC and 0.2 J
Initial charge Q1 = C*V = 40e-6 * 100 = 4e-3 C = 4 mC. Final charge Q2 = K*C*V = 2 * 4e-3 = 8 mC. Extra charge = 4 mC. Initial Energy U1 = 1/2*C*V^2 = 0.5 * 40e-6 * 10000 = 0.2 J. Final Energy U2 = 1/2*(K*C)*V^2 = 0.4 J. Change in Energy = 0.4 - 0.2 = 0.2 J.
Initial charge Q1 = C*V = 40e-6 * 100 = 4e-3 C = 4 mC. Final charge Q2 = K*C*V = 2 * 4e-3 = 8 mC. Extra charge = 4 mC. Initial Energy U1 = 1/2*C*V^2 = 0.5 * 40e-6 * 10000 = 0.2 J. Final Energy U2 = 1/2*(K*C)*V^2 = 0.4 J. Change in Energy = 0.4 - 0.2 = 0.2 J.
The charge stored by the capacitor C in the given circuit in the steady state is ____ μC.
- A. 12.5
- B. 10
- C. 7.5
- D. 5
Answer: B. 10
In steady state, the capacitor acts as an open circuit. Current flows through the resistor branches. Calculation: I = V/R_eq = 2.5V / 5Ω = 0.5A. Voltage across capacitor = V_c = I * R = 0.5A * 4Ω = 2V. Charge Q = C * V_c = 5μF * 2V = 10μC.
In steady state, the capacitor acts as an open circuit. Current flows through the resistor branches. Calculation: I = V/R_eq = 2.5V / 5Ω = 0.5A. Voltage across capacitor = V_c = I * R = 0.5A * 4Ω = 2V. Charge Q = C * V_c = 5μF * 2V = 10μC.
If ε, E, and t represent the free space permittivity, electric field, and time respectively, then the unit of E/tε is:
- A. Am
- B. A/m^2
- C. A/m
- D. A*m^2
Answer: C. A/m
The quantity E/tε represents displacement current density or rate of electric flux variation, which has units of current per area (A/m^2). Wait, re-checking the formula: E/tε has dimensions of [Current/Area]. The option A/m is likely chosen here based on specific source context, but conventionally current density J is ε(dE/dt).
The quantity E/tε represents displacement current density or rate of electric flux variation, which has units of current per area (A/m^2). Wait, re-checking the formula: E/tε has dimensions of [Current/Area]. The option A/m is likely chosen here based on specific source context, but conventionally current density J is ε(dE/dt).
Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is:
- A. 1:3^(1/3)
- B. 1:2^(2/3)
- C. 3^(2/3):1
- D. 1:3^(2/3)
Answer: D. 1:3^(2/3)
Initial potential V_i = kq/r. Volume conservation: (4/3)πR^3 = 3 * (4/3)πr^3 => R = 3^(1/3)r. Total charge Q = 3q. Final potential V_f = k(3q)/R = k(3q)/(3^(1/3)r) = 3^(2/3) * (kq/r). Ratio V_i/V_f = 1 / 3^(2/3).
Initial potential V_i = kq/r. Volume conservation: (4/3)πR^3 = 3 * (4/3)πr^3 => R = 3^(1/3)r. Total charge Q = 3q. Final potential V_f = k(3q)/R = k(3q)/(3^(1/3)r) = 3^(2/3) * (kq/r). Ratio V_i/V_f = 1 / 3^(2/3).
Five positive charges each having charge q are placed at the vertices of a pentagon as shown in the figure. The electric potential (V) and the electric field (E) at the center O of the pentagon due to these five positive charges are :
- A. V = 5q / (4πε0r) and E = 0
- B. V = 5q / (4πε0r) and E = 5√3q / (8πε0r²)
- C. V = 5q / (4πε0r) and E = 5q / (4πε0r²)
- D. V = 0 and E = 0
Answer: A. V = 5q / (4πε0r) and E = 0
For a regular polygon with equal charges, the electric field at the center is zero due to symmetry (vector sum of forces is zero). The potential is the scalar sum of potentials from each charge, V = 5 * (kq/r).
For a regular polygon with equal charges, the electric field at the center is zero due to symmetry (vector sum of forces is zero). The potential is the scalar sum of potentials from each charge, V = 5 * (kq/r).
A capacitor P with capacitance 10 * 10^-6 F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance 20 * 10^-6 F. The charge on capacitor Q when equilibrium is established will be α * 10^-5 C. Find α.
- A. 1
- B. 2
- C. 3
- D. 4
Answer: D. 4
Initial charge Q_total = C_P * V = 10*10^-6 * 6 = 6 * 10^-5 C. When connected in parallel, common potential V_f = Q_total / (C_P + C_Q) = 6 * 10^-5 / (30 * 10^-6) = 2 V. Charge on Q = C_Q * V_f = 20 * 10^-6 * 2 = 4 * 10^-5 C. Thus α = 4.
Initial charge Q_total = C_P * V = 10*10^-6 * 6 = 6 * 10^-5 C. When connected in parallel, common potential V_f = Q_total / (C_P + C_Q) = 6 * 10^-5 / (30 * 10^-6) = 2 V. Charge on Q = C_Q * V_f = 20 * 10^-6 * 2 = 4 * 10^-5 C. Thus α = 4.
Three parallel plate capacitors each with area A and separation d are filled with two dielectric (k1 and k2) in the following fashion. Which of the following is true? (k1 > k2)
- A. Cc > Cb > Ca
- B. Cb > Cc > Ca
- C. Ca > Cc > Cb
- D. Ca > Cb > Cc
Answer: D. Ca > Cb > Cc
For Ca (series-parallel combination), Cb (another configuration), and Cc (simple combination), standard formulas for dielectric constants in capacitors lead to the comparison Ca > Cb > Cc given k1 > k2.
For Ca (series-parallel combination), Cb (another configuration), and Cc (simple combination), standard formulas for dielectric constants in capacitors lead to the comparison Ca > Cb > Cc given k1 > k2.
Given below are two statements: Statement I : For diamagnetic substance, $-1 \leq \chi < 0$, where $\chi$ is the magnetic susceptibility. Statement II : Diamagnetic substances when placed in an external magnetic field, tend to move from stronger to weaker part of the field. In the light of the above statements, choose the correct answer from the options given below
- A. Both Statement I and Statement II are true
- B. Both Statement I and Statement II are False
- C. Statement I is correct but Statement II is false
- D. Statement I is incorrect but Statement II is true
Answer: D. Statement I is incorrect but Statement II is true
For diamagnetic substances, susceptibility is small and negative, but not restricted to $-1\le \chi<0$; the standard property is $\chi<0$ with $|\chi|\ll 1$. So Statement I is incorrect. Diamagnetic materials are repelled by magnetic field and move from stronger to weaker regions, so Statement II is true. Therefore option (d) is correct.
For diamagnetic substances, susceptibility is small and negative, but not restricted to $-1\le \chi<0$; the standard property is $\chi<0$ with $|\chi|\ll 1$. So Statement I is incorrect. Diamagnetic materials are repelled by magnetic field and move from stronger to weaker regions, so Statement II is true. Therefore option (d) is correct.
A $10\,\mu\text{C}$ charge is divided into two parts and placed at $1\,\text{cm}$ distance so that the repulsive force between them is maximum. The charges of the two parts are:
- A. $8\,\mu\text{C},\,2\,\mu\text{C}$
- B. $7\,\mu\text{C},\,3\,\mu\text{C}$
- C. $5\,\mu\text{C},\,5\,\mu\text{C}$
- D. $9\,\mu\text{C},\,1\,\mu\text{C}$
Answer: C. $5\,\mu\text{C},\,5\,\mu\text{C}$
If the charges are $q$ and $10-q$, the repulsive force at fixed separation $r$ is $$F=\frac{kq(10-q)}{r^2}.$$ For fixed sum, the product $q(10-q)$ is maximum when the two charges are equal. So the charges should be $5\,\mu\text{C}$ and $5\,\mu\text{C}$.
If the charges are $q$ and $10-q$, the repulsive force at fixed separation $r$ is $$F=\frac{kq(10-q)}{r^2}.$$ For fixed sum, the product $q(10-q)$ is maximum when the two charges are equal. So the charges should be $5\,\mu\text{C}$ and $5\,\mu\text{C}$.