NCERT Exemplar Solutions Class 12 Physics Electric Charges and Fields — Multiple Choice Questions Q1–13

NCERT Exemplar Problems, Class 12 Physics — Electric Charges and Fields, MCQ I and MCQ II. All 13 questions solved.

The book numbers these questions 1.1 to 1.13; here they are Questions 1 to 13. Questions 1 to 7 are MCQ I (exactly one option is correct) and Questions 8 to 13 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Coulomb’s law and superposition. The force on a charge is the vector sum of the forces from every other charge, and like charges repel.
• Gauss’s law. $\displaystyle\oint_S \mathbf{E}\cdot d\mathbf{S} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$. The left side contains the field of all charges; the right side counts only the charge inside $S$.
• Conductors in electrostatics. The field inside is zero, and field lines meet the surface at right angles.
• Symmetry. A component of the field that would have to point one way as often as the other cancels, which is how Questions 6, 7, 12 and 13 are settled without integrals.

Key insight. Flux depends on the enclosed charge and on nothing else: not the shape of the surface, not where the charge sits inside it, not the charges outside. Two printed answers on this page go wrong precisely by forgetting that a surface can enclose charges that cancel (Question 8) or by misjudging a stability question (Question 13).

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

In the figure, two positive charges $q_2$ and $q_3$ are fixed along the $y$-axis, $q_2$ above the origin and $q_3$ below it. Together they exert a net electric force in the $+x$ direction on a charge $q_1$ fixed on the $x$-axis, on the negative-$x$ side of the origin. If a positive charge $Q$ is now added at $(x, 0)$, on the positive-$x$ side, the force on $q_1$

Solution. First find the sign of $q_1$. The positive charges $q_2$ and $q_3$ lie on the $y$-axis, to the right of $q_1$. If $q_1$ were positive they would push it to the left; the net force is towards $+x$, so they are pulling it, and $q_1$ must be negative.

Now add $Q > 0$ on the far right. A positive charge attracts the negative $q_1$, so the new force on $q_1$ points towards $Q$, again along $+x$. Forces simply add by superposition. The force from $Q$ does not alter the forces from $q_2$ and $q_3$; charges do not “intersect” or shield one another, which disposes of (d).

So the net force keeps its direction and grows.

Question 2

A point positive charge is brought near an isolated conducting sphere. The electric field is best given by which of the four figures?

Each of the four figures shows the point charge $+q$ on the left of a grey sphere with field lines drawn between them. In (i) and (ii) the lines from $+q$ terminate on the near face of the sphere, and further lines appear on the far side; the figures differ in the arrow directions on the far side. In (iii) and (iv) the lines are drawn running through the sphere’s outline.

Solution. Apply three facts about a conductor in electrostatic equilibrium.

1. The field inside the conductor is zero, so no line may enter the sphere’s interior. Lines must end on the surface. That rules out (iii) and (iv), where lines pass straight through the sphere.
2. Field lines meet a conducting surface at right angles, and they start on positive charge and end on negative charge.
3. The charge $+q$ induces negative charge on the near side of the sphere, where the lines coming from $+q$ end. The sphere is isolated and neutral, so an equal positive charge appears on the far side, and lines start there and run outwards.

Figure (i) has exactly this pattern: lines from $+q$ end on the near side, and fresh lines leave the far side directed away from the sphere. Figure (ii) draws the far-side arrows pointing into the sphere, which would mean the far side is negative and the sphere carries a net negative charge, impossible for a neutral isolated sphere.

Question 3

The electric flux through the surface

Four closed surfaces are drawn. (i) is an irregular bulb-shaped surface with $+q$ inside it, (ii) is a circle with $+q$ off-centre inside, (iii) is an ellipse with $+q$ near its centre, and (iv) is a large, strongly indented irregular surface with $+q$ inside. All four are closed surfaces S enclosing the same charge $+q$.

Solution. By Gauss’s law the flux through any closed surface is $q/\varepsilon_0$, where $q$ is the charge inside. Each figure encloses the same charge $+q$, so each flux equals $q/\varepsilon_0$.

The shape of the surface changes where the field lines cross it, and how steeply, but every line that leaves the charge must pass out through the surface, so the total count never changes. That is why a crumpled surface and a neat sphere give the same answer.

Question 4

Five charges $q_1, q_2, q_3, q_4$ and $q_5$ are fixed at the positions shown. $S$ is a Gaussian surface. In the figure the curved surface $S$ encloses $q_2$ and $q_4$ only, while $q_1$, $q_3$ and $q_5$ lie outside it. Gauss’s law is $\oint_S \mathbf{E}\cdot d\mathbf{s} = q/\varepsilon_0$. Which statement is correct?

Solution. $\mathbf{E}$ in the integrand is the total field at each point of the surface, produced by every charge in the region, inside or outside $S$. The outside charges $q_1, q_3, q_5$ do contribute to $\mathbf{E}$.

Their contribution to the integral, however, vanishes: each outside charge sends as many field lines into the surface as out of it. That is why the right side counts only the enclosed charge, $q = q_2 + q_4$.

So $\mathbf{E}$ comes from all five charges, and $q$ from $q_2$ and $q_4$ only.

Question 5

The figure shows electric field lines in which an electric dipole $\mathbf{p}$ is placed. The field lines are five nearly horizontal lines, all pointing to the right, which spread apart as they go from left to right, so they are closer together on the left. The dipole lies along a line with $-q$ at its left end and $+q$ at its right end, so $\mathbf{p}$ points to the right.

Solution. The field is not uniform, so the two ends feel unequal forces and a net force remains. The field points right everywhere, so:

• $+q$ at the right end is pushed to the right with force $qE_{\text{right}}$;
• $-q$ at the left end is pushed to the left with force $qE_{\text{left}}$.

Field-line density measures field strength. The lines are closer together on the left, so $E_{\text{left}} > E_{\text{right}}$, and the leftward force on $-q$ is the larger one. The net force is therefore towards the left.

Check with the formula: $F_x = p\,\dfrac{dE}{dx}$. Here $E$ decreases to the right, so $dE/dx < 0$ and $F_x < 0$.

Question 6

A point charge $+q$ is placed at a distance $d$ from an isolated conducting plane. The field at a point P on the other side of the plane is

Solution. The plane is an isolated conductor, so it is neutral overall. The charge $+q$ induces negative charge on the near face and an equal positive charge on the far face. A point P just beyond the plane lies outside the conductor, and the field there is due to the far-face positive charge together with everything else.

Two points settle the direction. The field at the surface of a conductor is always normal to the surface, which excludes the radial options (c) and (d). And the far face carries net positive charge, so the lines start there and point away from the plane.

Question 7

A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed

Solution. The “diameter” is the diameter of the flat circular base. Take a point on it, away from the centre. Every element of the curved surface has a mirror partner on the other side of the plane containing the point and the hemisphere’s axis. The two contributions have equal components along the axis and opposite components sideways, so the components lying in the plane of the base cancel in pairs, while the components along the axis of the hemisphere add.

Only the component perpendicular to the base survives. The field at that point is therefore perpendicular to the diameter.

Question 8

If $\oint_S \mathbf{E}\cdot d\mathbf{S} = 0$ over a surface, then

Solution. Zero net flux says only that the net enclosed charge is zero. Nothing more follows.

• (a) is false. A surface can enclose nothing and still sit in a strong field from outside charges.
• (b) is false. Zero net flux says nothing about uniformity; a dipole inside the surface gives zero flux and a field that varies strongly.
• (c) is true. Net flux is lines out minus lines in, and that is zero.
• (d) is false. A charge $+q$ and a charge $-q$ can both lie inside the surface; the net flux is $(q – q)/\varepsilon_0 = 0$, yet the charges are not outside.

Printed key error. The key prints (c), (d). Option (d) says charges must necessarily be outside, and the dipole example refutes it. Only (c) is correct.

Question 9

The electric field at a point is

Solution. Away from charges the field is a smooth function of position, so it is continuous at any point where no charge sits. That gives (b).

At a point charge the field $\dfrac{q}{4\pi\varepsilon_0 r^2}$ diverges as $r \to 0$ and points in opposite directions on opposite sides, so it is discontinuous there. The sign of the charge is irrelevant, which kills (c). A charged surface likewise produces a jump of $\sigma/\varepsilon_0$ across it. “Always continuous” in (a) is false for the same reason.

Question 10

If there were only one type of charge in the universe, then

Solution. Gauss’s law does not need two kinds of charge; it only relates flux to the enclosed charge.

• (a) is false: a surface that encloses no charge has zero flux, even if charges of the single type lie elsewhere.
• (b) is true: charge outside contributes equal entering and leaving flux.
• (c) is false: the flux integral is defined for any field.
• (d) is true: it is Gauss’s law itself.

Question 11

Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region

Solution. A neutral distribution has no monopole term, whose field would go as $1/r^2$. The leading term at large $r$ is the dipole term, and the dipole field falls as $1/r^3$. So (c) is correct.

(a) is false: a dipole is neutral and still has a field. (b) is false: the dipole term dominates but is not the only one, since quadrupole and higher terms also contribute, falling off faster. (d) is true because the electrostatic field is conservative, so the work around any closed path is zero.

Question 12

Refer to the arrangement of charges in the figure and a Gaussian surface of radius $R$ with $Q$ at the centre. The charge $5Q$ lies on a horizontal line through the centre, outside the sphere at a distance $2R$ from the centre. The charge $-2Q$ lies inside the sphere, directly below the centre at a distance $R/2$. Then

Solution. Charge inside the sphere: $Q + (-2Q) = -Q$. The charge $5Q$ is outside.

• (a) True: $\Phi = \dfrac{-Q}{\varepsilon_0}$.
• (b) False. That formula would hold only if the field were that of a point charge $-Q$ at the centre. The charges are not symmetrically placed, and $5Q$ also contributes, so the field on the surface is not uniform.
• (c) True: a charge outside the surface gives zero net flux through it.
• (d) False: $-2Q$ is off-centre, so its distance from different points of the surface varies and so does its field.

Question 13

A positive charge $Q$ is uniformly distributed along a circular ring of radius $R$. A small test charge $q$ is placed at the centre of the ring. Then

Solution. The field at the centre is zero by symmetry, so the centre is an equilibrium point for any $q$. The question is stability, and the in-plane and axial directions behave oppositely.

Along the axis, at height $z$ above the centre, the field is $$E_z = \frac{kQz}{(R^2+z^2)^{3/2}} \approx \frac{kQz}{R^3} \qquad (k = 1/4\pi\varepsilon_0)$$ directed away from the ring on either side. A negative $q$ feels $F_z = -\dfrac{kQq_{\text{mag}}}{R^3}z$, a restoring force proportional to $z$, so it performs SHM. That is (c). A positive $q$ is pushed further away along the axis: unstable along the axis.

In the plane, the electrostatic potential $V$ obeys Laplace’s equation, $\nabla^2 V = 0$, in the empty region near the centre. Along the axis $V = kQ/\sqrt{R^2+z^2}$ has a maximum at the centre. Since the three second derivatives sum to zero, the in-plane second derivatives must be positive overall, so the centre is a minimum of $V$ in the plane. A positive $q$ sits at a minimum of its potential energy, so it is stable in the plane and is pushed back. That is (a).

For a negative $q$ the energy picture inverts: the centre is a maximum of $qV$ in the plane, so a displaced $q < 0$ is pushed outward and keeps going until it reaches the ring. That is (b).

Numerical check: $V(0.1R$ in the plane$)/V(0) = 1.0025 > 1$, so the potential rises away from the centre in the plane (minimum at the centre); $V(0.1R$ on the axis$)/V(0) = 0.995 < 1$, so it falls along the axis (maximum).

(d) says a positive $q$ is unstable within the plane. That contradicts (a) and the working above.

Printed key error. The key prints (a), (b), (c) and (d). Option (d) cannot be right alongside (a): a charge cannot be both pushed back and unstable in the same plane. The correct answer is (a), (b), (c).

Common mistakes

• Trusting the printed key on Question 8. It adds (d), but zero net flux does not mean all charges are outside; a $+q$ and a $-q$ both inside also give zero.
• Trusting the printed key on Question 13. It adds (d), yet a positive charge at the centre of a ring is stable in the plane and unstable along the axis.
• Putting only the enclosed charges into $\mathbf{E}$ in Question 4. The field in the integrand is the total field from every charge; only the right-hand side counts the enclosed charge.
• Treating flux as shape-dependent in Question 3. The flux through any closed surface depends only on the enclosed charge.
• Judging the force on a dipole from the field direction alone in Question 5. In a uniform field the net force is zero; here the field is non-uniform, so the stronger end wins.
• Letting field lines enter a conductor in Question 2. Lines end on the surface, since the field inside is zero.
• Reading $E \propto 1/r^2$ for a neutral distribution in Question 11. With no net charge the leading term is the dipole field, $\propto 1/r^3$.
• Using a point-charge formula for the field on a surface in Question 12. Gauss’s law gives the flux, not the field, unless the symmetry is perfect.

Practise next

• Exemplar Electric Charges and Fields: Very Short, Short and Long Answer Questions — where the same Gauss’s law and symmetry arguments are applied to conductors, cubes, the shell model and oscillation of charges.
• NCERT textbook Chapter 1 — Electric Charges and Fields for the theory of dipoles, Gauss’s law and field of a ring.
• Exemplar Electrostatic Potential and Capacitance — the next unit, which recasts the force and field ideas here in terms of potential, including the Laplace-equation reasoning behind Question 13.