Every Book Back multiple-choice question from Electrostatics (12th Standard Physics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
Two identical point charges of magnitude –q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?
- A. A1 and A2
- B. B1 and B2Correct
- C. both directions
- D. No stable
Explanation. Stability is determined by the presence of a restoring force that returns the charge to its equilibrium position when displaced along specific axes.
Q2
Which charge configuration produces a uniform electric field?
- A. point charge
- B. uniformly charged infinite line
- C. uniformly charged infinite planeCorrect
- D. uniformly charged spherical shell
Explanation. An infinite plane of charge creates an electric field where the magnitude is constant at any distance from the plane, ensuring a uniform field configuration.
Q3
What is the ratio of the charges \(q_1 / q_2\) for the following electric field line pattern?
- A. \(1/5\)
- B. \(25/11\)
- C. \(5\)
- D. \(11/25\)Correct
Explanation. The magnitude of a point charge is directly proportional to the number of electric field lines that originate from or terminate on that charge.
Q4
An electric dipole is placed at an alignment angle of \(30^\circ\) with an electric field of \(2 \times 10^5 \text{ N C}^{-1}\). It experiences a torque equal to \(8 \text{ N m}\). The charge on the dipole if the dipole length is \(1 \text{ cm}\) is
- A. \(4 \text{ mC}\)
- B. \(8 \text{ mC}\)Correct
- C. \(5 \text{ mC}\)
- D. \(7 \text{ mC}\)
Explanation. Torque is calculated using the relation \(\tau = pE \sin\theta\), where \(p\) is the dipole moment (\(q \times 2a\)), allowing us to solve for the charge.
Q5
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order.
- A. \(D < C < B < A\)Correct
- B. \(A < B = C < D\)
- C. \(C < A = B < D\)
- D. \(D > C > B > A\)
Explanation. According to Gauss Law, the total electric flux depends only on the net algebraic sum of charges enclosed by the Gaussian surface.
Q6
The total electric flux for the following closed surface which is kept inside water
- A. \(80q / \epsilon_0\)
- B. \(q / 40\epsilon_0\)Correct
- C. \(q / 80\epsilon_0\)
- D. \(q / 160\epsilon_0\)
Explanation. The total electric flux is the net enclosed charge divided by the permittivity of the medium, which is the product of vacuum permittivity and relative permittivity.
Q7
Two identical conducting balls having positive charges \(q_1\) and \(q_2\) are separated by a centre to centre distance \(r\). If they are made to touch each other and then separated to the same distance, the force between them will be
- A. less than before
- B. same as before
- C. more than beforeCorrect
- D. zero
Explanation. Touching identical conductors redistributes total charge equally. The resulting force is optimized because the product of equalized charges is always greater than the initial product of unequal charges.
Q8
Rank the electrostatic potential energies for the given system of charges in increasing order.
- A. \(1 = 4 < 2 < 3\)Correct
- B. \(2 = 4 < 3 < 1\)
- C. \(2 = 3 < 1 < 4\)
- D. \(3 < 1 < 2 < 4\)
Explanation. Electrostatic potential energy is calculated using the formula \(U = k q_1 q_2 / r\), considering both the magnitude and signs of the charges.
Q9
An electric field \(\vec{E} = 10x \hat{i}\) exists in a certain region of space. Then the potential difference \(V = V_o - V_A\), where \(V_o\) is the potential at the origin and \(V_A\) is the potential at \(x = 2 \text{ m}\) is:
- A. \(10 \text{ V}\)
- B. \(-20 \text{ V}\)
- C. \(+20 \text{ V}\)Correct
- D. \(-10 \text{ V}\)
Explanation. The potential difference is calculated by integrating the electric field expression over the distance between the points, using the relation \(dV = -\vec{E} \cdot d\vec{r}\).
Q10
A thin conducting spherical shell of radius \(R\) has a charge \(Q\) which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is
- A. (a)
- B. (b)Correct
- C. (c)
- D. (d)
Explanation. Potential inside and on the surface of a conducting shell is constant, while outside it follows an inverse relationship with the distance from the center.
Q11
Two points A and B are maintained at a potential of \(7 \text{ V}\) and \(-4 \text{ V}\) respectively. The work done in moving \(50\) electrons from A to B is
- A. \(8.80 \times 10^{-17} \text{ J}\)Correct
- B. \(-8.80 \times 10^{-17} \text{ J}\)
- C. \(4.40 \times 10^{-17} \text{ J}\)
- D. \(5.80 \times 10^{-17} \text{ J}\)
Explanation. Work done is the product of the total charge of the electrons and the potential difference between the destination and the starting point.
Q12
If voltage applied on a capacitor is increased from \(V\) to \(2V\), choose the correct conclusion.
- A. \(Q\) remains the same, \(C\) is doubled
- B. \(Q\) is doubled, \(C\) doubled
- C. \(C\) remains same, \(Q\) doubledCorrect
- D. Both \(Q\) and \(C\) remain same
Explanation. Capacitance is independent of the applied voltage; therefore, increasing voltage causes the stored charge to increase proportionally according to \(Q = CV\).
Q13
A parallel plate capacitor stores a charge \(Q\) at a voltage \(V\). Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
- A. Capacitance
- B. Charge
- C. Voltage
- D. Energy densityCorrect
Explanation. Doubling both area and distance keeps capacitance constant, but since energy density depends inversely on the square of distance at constant voltage, it changes value.
Q14
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between the points A and C is
- A. \(1 \mu\text{F}\)
- B. \(2 \mu\text{F}\)Correct
- C. \(3 \mu\text{F}\)
- D. \(14 \mu\text{F}\)
Explanation. Equivalent capacitance is found by simplifying the circuit through the combination rules for capacitors connected in series and parallel branches.
Q15
Two metallic spheres of radii \(1 \text{ cm}\) and \(3 \text{ cm}\) are given charges of \(-1 \times 10^{-2} \text{ C}\) and \(5 \times 10^{-2} \text{ C}\) respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
- A. \(3 \times 10^{-2} \text{ C}\)Correct
- B. \(4 \times 10^{-2} \text{ C}\)
- C. \(1 \times 10^{-2} \text{ C}\)
- D. \(2 \times 10^{-2} \text{ C}\)
Explanation. Connecting spheres forces them to a common potential, redistributing the total net charge such that each sphere's share is proportional to its radius.
About these Electrostatics questions
These are the Book Back multiple-choice questions for Electrostatics from the Tamil Nadu State Board (Samacheer Kalvi) 12th Standard Physics syllabus. Each question shows the correct option and an original, step-by-step explanation so you understand the method, not just the answer. Use the answer key above to jump to any question, then take the practice test to check yourself under exam-like conditions.
Frequently asked questions
How many MCQs are there in Electrostatics?
This chapter has 15 book-back multiple-choice questions, each with the correct answer and a step-by-step explanation.
Are these 12th Standard Physics MCQs free to practise online?
Yes. Every question, answer and explanation here is free, and you can also take them as a timed practice test.
Where can I find the Electrostatics book-back answers?
The correct option for each question is highlighted on this page with a worked explanation, plus a quick answer-key summary at the top.
