Prepare for your CBSE Class 12 Physics board examination with Electric Charges and Fields Class 12 Board Paper MCQ. This collection includes important multiple-choice questions based on previous CBSE board examination papers and sample papers. These questions cover important topics such as electric charges, Coulomb’s law, electric field, electric dipole, electric flux, and Gauss’s law. Practising these MCQs will help students strengthen their conceptual understanding, revise important formulas, and improve their preparation for the Class 12 Physics board examination.
Electric Charges and Fields Class 12 Board Paper MCQ
Q. In a printer, neutral ink droplets are passed through an electron beam and some electrons are captured by the ink droplets and become charged. The net charge on an ink droplet is found to be − 1 x 10−12 C. The number of electrons captured by this ink droplet is: (Board Paper Comp 2026)
(A) 1·6 x 104
(B) 108
(C) 6·25 x 106
(D) 8 x 10
Answer (C) 6.25 × 10⁶ electrons are captured by the ink droplet.
Explanation:
According to the quantisation of charge, $q = ne$
Therefore,
$n=\frac{q}{e}$
$n=\frac{1\times10^{-12}}{1.6\times10^{-19}}$
$=6.25\times10^6$
A negative charge indicates that the droplet has gained electrons.
Hence, the number of electrons captured is $6.25\times10^6$
Q. Which of the following is not correct about the electrostatic force between two charges? (Board Paper Comp 2026)
(A) The magnitude of the force is directly proportional to the magnitudes of the two charges.
(B) It is inversely proportional to the square of the distance between the two charges.
(C) It depends on the permittivity of the medium in which the charges are placed.
(D) It is inversely proportional to the distance between the two charges.
Answer (D) It is inversely proportional to the distance between the two charges.
Explanation:
According to Coulomb’s law,
$F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}$
The electrostatic force is inversely proportional to the square of the distance between the charges.
$F\propto \frac{1}{r^2}$
Therefore, statement (D) is incorrect.
Q. How many protons are needed to make 1 nC charge? (Board Paper 2026)
(A) 1.60 x 10+28
(B) 1.60 x 10+10
(C) 3.1 x 109
(D) 6.25 x 109
Answer (D) 6.25 × 10⁹
Explanation:
Given,
$q=1,nC=10^{-9}C$
Using
$q=ne$
$n=\frac{10^{-9}}{1.6\times10^{-19}}$
$=6.25\times10^9$
Hence,
$6.25\times10^9$
protons are required.
Q. A particle of mass $m$ and charge $q$ starts from rest and moves in an electric field $\vec{E} = E_0\hat{i}$. After travelling a distance $x$ in the field along the x-axis, the kinetic energy of the particle will be: (Board Paper 2026)
(A) $\dfrac{qE_0x}{2}$
(B) $qE_0x$
(C) $q^2E_0x$
(D) $q^2E_0^2x^2$
Answer (B) $qE_0x$
Explanation:
The force acting on the particle is
$F=qE_0$
Work done by the electric field is
$W=Fx=qE_0x$
According to the work-energy theorem,
$W=\Delta K$
Hence,
$K=qE_0x$
Q. Two small identical metallic balls having charges q and – 2q are kept far at a separation r. They are brought in contact and then separated at distance r/2. Compared to the initial force F, they will now : (Board Paper 2026)
(A) attract with a force F/2
(B) repel with a force F/2
(C) repel with a force F
(D) attract with a force F
Answer (B) Repel with a force F/2.
Explanation:
Initial charge:
$q+(-2q)=-q$
After contact, each ball gets
$-\frac{q}{2}$
Initial force:
$F=\frac{k(2q^2)}{r^2}$
New force:
$F’=\frac{k\left(\frac{q}{2}\right)^2}{(r/2)^2}$
$=\frac{kq^2}{r^2}$
Therefore,
$F’=\frac{F}{2}$
Since both balls have negative charges after contact, the force is repulsive.
Q. Which of the following statements about electric field E is incorrect ?(Board Paper Comp 2025)
(A) The magnitude of the electric field due to a point charge is proportional to 1/r2, where r is the distance of the point from the charge.
(B) The magnitude of the electric field is zero at all points in a region where the potential is zero.
(C) Electric field is a vector quantity.
(D) The magnitude of the electric field at a point is a measure of the force experienced by a unit charge at that point.
Answer (B) The magnitude of the electric field is zero at all points in a region where the potential is zero.
Explanation:
Zero electric potential at a point does not necessarily mean that the electric field is zero at that point.
A region may have zero potential but still possess an electric field.
Therefore statement (B) is incorrect.
Q. A small metallic sphere S of charge +q is placed exactly at a point midway between two-point charges A (+Q) and B (+Q) where Q >> q. If the sphere is slightly displaced towards charge A and released, then:(Board Paper Comp 2025)
(A) it will move further towards A
(B) it will move towards B
(C) it will oscillate about its original position
(D) it will not move at all
Answer (B) It will move towards B.
Explanation:
At the midpoint, forces due to the two equal charges cancel.
When displaced slightly towards A, the repulsive force due to A becomes larger than the repulsive force due to B.
Hence the net force acts away from A and towards B.
Therefore the sphere moves towards B.
Q. A charge Q is enclosed by a spherical Gaussian surface of radius R. If the radius is doubled, then the total electric flux through the surface :(Board Paper Comp 2025)
(A) becomes four times
(B) remains the same
(C) becomes half
(D) becomes twice
Answer (B) Remains the same.
Explanation:
According to Gauss’s law,
$\Phi=\frac{Q}{\varepsilon_0}$
Electric flux depends only on the enclosed charge and not on the size of the Gaussian surface.
Hence the flux remains unchanged.
Q. A charge of -1 𝝁C on a body represents (Board Paper 2025)
(A) loss of 6.25 x 1012 electrons by the body.
(B) gain of 6.25 x 1012 electrons by the body.
(C) gain of 1.6 x 1013 electrons by the body.
(D) loss of 1.6 x 1013 electrons by the body.
Answer (B) gain of 6.25 x 1012 electrons by the body.
Explanation:
$q=-1\times10^{-6}C$
Using
$n=\frac{q}{e}$
$n=\frac{10^{-6}}{1.6\times10^{-19}}$
$=6.25\times10^{12}$
Negative charge means gain of electrons.
Q. Two charges —q each are placed at the vertices A and B of an equilateral triangle ABC. If M is the mid-point of AB, the net electric field at C will point along (Board Paper 2025)
(A) CA
(B) CB
(C) MC
(D) CM
Answer (D) CM
Explanation:
Electric field due to a negative charge is directed towards the charge.
At point C, the electric fields due to charges at A and B are equal.
The electric field due to a negative charge is directed towards the charge. The resultant electric field at C is along CM.
Hence the net electric field points towards M.
Q. Which one of the following statements is correct? (Board Paper 2025)
Electric field due to static charges is
(A) conservative and field lines do not form closed loops.
(B) conservative and field lines form closed loops.
(C) non-conservative and field lines do not form closed loops.
(D) non-conservative and field lines form closed loops.
Answer (A) conservative and field lines do not form closed loops.
Explanation:
Electrostatic field is conservative. Electrostatic field lines never form closed loops. Therefore option (A) is correct.
Q. A body acquires charge 8:0 x 10-12 C. The mass of the body : (Board Paper 2025)
(A) increases by 4.5 x 10-7 kg
(B) decreases by 1.0 x 10-6 kg
(C) decreases by 4.55 x 10-23 kg
(D) increases by 9.1 x 10-23 kg
Answer (C) decreases by 4.55 x 10-23 kg
Explanation:
A body becomes positively charged by losing electrons. The number of electrons lost is
$n=\frac{q}{e}$
$n=\frac{8.0\times10^{-12}}{1.6\times10^{-19}}$
$=5\times10^7$
Mass of one electron is
$9.1\times10^{-31},kg$
Therefore, decrease in mass is
$\Delta m=nm_e$
$\Delta m=5\times10^7\times9.1\times10^{-31}$
$\Delta m=4.55\times10^{-23},kg$
Hence, the mass of the body decreases by
$4.55\times10^{-23},kg$
Q. Two horizontal plates, separated by 1 cm, are arranged one above the other. A particle of mass 5 mg and charge 2 nC is released in air between the plates. The potential difference that should be applied to the plates so that the particle remains suspended between them, is : (Board Paper 2025)
(A) 250V
(B) 200V
(C) 100V
(D) 50V
Answer (A) 250 V
Explanation:
For the particle to remain suspended, the electric force must balance its weight.
$qE=mg$
Therefore,
$E=\frac{mg}{q}$
Substituting,
$m=5\times10^{-6},kg$
$q=2\times10^{-9},C$
$g=9.8,m,s^{-2}$
$E=\frac{5\times10^{-6}\times9.8}{2\times10^{-9}}$
$=2.45\times10^4,N/C$
Using
$E=\frac{V}{d}$
$V=Ed$
$V=2.45\times10^4\times10^{-2}$
$V\approx245V\approx250V$
Hence, the required potential difference is 250 V.
Q. A thin plastic rod is bent into a circular ring of radius R. It is uniformly charged with charge density λ. The magnitude of the electric field at its centre is : (Board Paper 2024)
(A) λ / 2ε0R
(B) Zero
(C) λ / 4πε0R
(D) λ / 4ε0R
Answer (B) Zero
Explanation:
The charge is uniformly distributed over the complete circular ring.
For every small charge element on the ring, there is an opposite charge element producing an equal electric field in the opposite direction at the centre.
Therefore, all electric field vectors cancel each other.
Hence, the net electric field at the centre is $E=0$
Q. Two charged particles P and Q, having the same charge but different masses mP and mQ, start from rest and travel equal distances in a uniform electric field E→ in time tP and tQ respectively. Neglecting the effect of gravity, the ratio (tP/tQ) is: (Board Paper 2024)
(A) mP/mQ
(B) mQ/mP
(C) √mP/mQ
(D) √mQ/mP
Answer (C) √mP/mQ
$\sqrt{\frac{m_P}{m_Q}}$
Explanation:
The force acting on each particle is
$F=qE$
Acceleration is
$a=\frac{qE}{m}$
Using the equation of motion,
$s=\frac{1}{2}at^2$
$t=\sqrt{\frac{2s}{a}}
Since
$a\propto \frac{1}{m}$
therefore,
$t\propto\sqrt{m}$
Hence,
$\frac{t_P}{t_Q}$
$=$
$\sqrt{\frac{m_P}{m_Q}}$
Q. Two identical small conducting balls B1 and B2, are given —7 pC and + 4 pC charges respectively. They are brought in contact with a third identical ball By and then separated. If the final charge on each ball is -2 pC, the initial charge on B3 was (Board Paper 2024)
(A) -2 pC
(B) -3 pC
(C) -5 pC
(D) -15 pC
Answer (B) −3 pC
Explanation:
According to the law of conservation of charge, total charge remains constant.
Initial charge:
$(-7)+(4)+q=-3+q$
Final charge:
$(-2)+(-2)+(-2)=-6,pC$
Therefore,
$-3+q=-6$
$q=-3,pC$
Hence, the initial charge on B3B_3B3 was
$-3,pC$
Q. When a glass rod is rubbed with a silk cloth, it acquires a positive charge of 4·8 10 -17 C. In the process, the number of electrons and protons transferred are respectively : (Board Paper 2024)
(A) 0, 300
(B) 0, 225
(C) 150, 150
(D) 300, 0
Answer (D) 300, 0
Explanation:
The charging by friction occurs due to transfer of electrons. Protons do not move from one body to another.
Number of electrons transferred:
$n=\frac{q}{e}$
$n=\frac{4.8\times10^{-17}}{1.6\times10^{-19}}$
$n=300$
Since the glass rod becomes positively charged, it loses 300 electrons.
No protons are transferred.
Hence, electrons transferred = 300 and protons transferred = 0.
Q. The electric flux through a Gaussian spherical surface enclosing a point charge q is Φ. If the charge is replaced by an electric dipole, magnitude of its dipole moment being 2qa, the flux through the surface will be : (Board Paper Comp 2023)
(a) 2 Φ
(b) Φ
(c) Φ/2
(d) Zero
Answer (D) Zero
Explanation:
According to Gauss’s law,
$\Phi=\frac{q}{\varepsilon_0}$
An electric dipole consists of equal and opposite charges.
Therefore, the total charge enclosed by the Gaussian surface is
$+q+(-q)=0$
Hence,
$\Phi=\frac{0}{\varepsilon_0}=0$
Therefore, the electric flux through the surface is zero.
Q. 109 electrons are transferred to a pith ball with charge 0·16 nC. Its charge now is : (Board Paper Comp 2023)
(a) Zero
(b) 3·2 x 10-10 C
(c) -1·6 x 10-9 C
(d) 3·2 x 10-10 C
Answer (A) Zero
Explanation:
Charge brought by 10910^9109 electrons is
$q=(10^9)(1.6\times10^{-19})$
$q=1.6\times10^{-10}C$
$q=0.16,nC$
The pith ball initially has
$+0.16,nC$
After gaining electrons,
$+0.16,nC-0.16,nC=0$
Thus, the pith ball becomes electrically neutral.
Q. Assertion (A) : The electrostatic field E→ is a conservative field.
Reason (R) : Line integral of electric field E→ around a closed path is non zero. (Board Paper Comp 2023)
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Both Assertion (A) and Reason (R) are false.
Answer (C) Assertion (A) is true, but Reason (R) is false.
Explanation:
The electrostatic field is a conservative field. For a conservative field, work done around a closed path is zero.
Therefore,
$\oint \vec{E}\cdot d\vec{l}=0$
The assertion is true, but the reason is false.
Q. The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N/C. From the same charged object the electric field of magnitude, 16 N/C will be at a distance of (Board Paper 2023)
(a) 1m
(b) 2m
(c) 3m
(d) 6m
Answer (C) 3 m
Explanation:
For a point charge,
$E\propto\frac{1}{r^2}$
Therefore,
$\frac{E_1}{E_2}=\frac{r_2^2}{r_1^2}$
$\frac{9}{16}=\frac{r^2}{4^2}$
$r^2=9$
$r=3,m$
Hence, the required distance is 3 m.
Q. Assertion (A) : Work done in moving a charge around a closed path, in an electric field is always zero.
Reason (R) : Electrostatic force is a conservative force. (Board Paper 2023)
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Both Assertion (A) and Reason (R) are false.
Answer (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion.
Explanation:
The electrostatic field is a conservative field. For a conservative force, the work done depends only on the initial and final positions and not on the path followed.
Therefore, for a closed path,
$\oint \vec{E} \cdot d\vec{l}=0$
Hence, the work done in moving a charge around a closed path is zero. Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.
Q. Which one of the following is not a scalar quantity ? (Board Paper 2023)
(a) Electric field
(b) Voltage
(c) Resistivity
(d) Power
Answer (a) Electric field
Explanation:
The electric field is a vector quantity because it has both magnitude and direction. The direction of electric field at a point is the direction of the force experienced by a positive test charge placed at that point.
Voltage, resistivity and power are scalar quantities.
Hence, electric field is not a scalar quantity.
Q. An electric dipole of length 2 cm is placed at an angle of 30° with an electric field 2 x 105 N/C. If the dipole experiences a torque of 8 x 10-3 Nm, the magnitude of either charge of the dipole, is (Board Paper 2023)
(A) 4 μC
(B) 7 μC
(C) 8 mC
(D) 2 mC
Answer (A) 4 μC4
Explanation:
According to NCERT,
$\tau = pE\sin\theta$
and
$p=q(2a)$
Given,
$\tau = 8\times10^{-3},Nm$
$E=2\times10^5,N/C$
$2a=2,cm=2\times10^{-2},m$
$\theta=30^\circ$
Therefore,
$q=\frac{\tau}{(2a)E\sin\theta}$
$q=\frac{8\times10^{-3}}{(2\times10^{-2})(2\times10^5)(1/2)}$
$q=4\times10^{-6},C$
$q=4,\mu C$
Hence, the charge on either end of the dipole is 4 μC.
Q. The charge on a body is 8 x 1012 C. It means that the body has : (Board Paper 2023)
(a) lost 8 x 1012 electrons
(b) gained 4 x 1010 electrons
(c) gained 2 x 108 electrons
(d) lost 5 x 107 electrons
Answer (d) lost 5 x 107 electrons
Explanation:
According to the quantisation of charge,
$q=ne$
Therefore,
$n=\frac{q}{e}$
$n=\frac{8\times10^{-12}}{1.6\times10^{-19}}$
$n=5\times10^7
A positive charge on a body indicates loss of electrons.
Hence, the body has lost
$5\times10^7$
electrons.
Q. If a charged hollow sphere and a solid sphere of aluminum and copper of equal radii are in electrostatic equilibrium, then which of the following statements is true? (Sample Paper 2026)
(A) Both the spheres are having equal charges.
(B) The hollow sphere will have more charge than solid sphere at its surface.
(C) The aluminum sphere will have more charge on its surface than copper sphere.
(D) If hollow sphere is also made up of aluminum then it will have more charge.
Answer (A) Both the spheres are having equal charges.
Explanation:
In electrostatic equilibrium, charge resides on the outer surface of a conductor. The distribution of charge does not depend upon whether the conductor is hollow or solid.
Since both spheres have equal radii and are conductors, the material or hollow-solid nature does not make one sphere hold more charge than the other.
Hence, option (A) is correct.
Q. Assertion (A): For three point charges to be in equilibrium, they must be collinear.
Reason(R): One of the three charges must have different polarity than rest of the two. (Sample Paper 2026)
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Both Assertion (A) and Reason (R) are false.
Answer (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion.
Explanation:
For equilibrium, the net force on each charge must be zero.
According to the principle of superposition, the forces acting on a charge must be equal and opposite. This condition is possible only when the three charges lie on the same straight line.
Also, one charge must be of opposite sign so that attractive and repulsive forces can balance each other.
Hence, both Assertion and Reason are true, and the Reason correctly explains the Assertion.
Q. An electric dipole placed in an electric field of intensity 2 × 105 N/C at an angle of
30° experiences a torque equal to 4 Nm. The charge on the dipole of dipole length 2 cm is (Sample Paper 2024)
(a) 7 µC
(b) 8 mC
(c) 2 mC
(d) 5 mC
Answer (d) 2 mC
Explanation:
Using NCERT relation,
$\tau=pE\sin\theta$
where
$p=q(2a)$
Given,
$\tau=4,Nm$
$E=2\times10^5,N/C$
$2a=2\times10^{-2},m$
$\sin30^\circ=\frac12$
Therefore,
$q=\frac{4}{(2\times10^{-2})(2\times10^{5})(1/2)}$
$q=2\times10^{-3},C$
$q=2,mC$
Hence, the required charge is 2 mC.
Q. Two point charges +8q and -2q are located at x=0 and x=L respectively. The point on x axis at which net electric field is zero due to these charges is- (Sample Paper 2022)
(i) 8L
(ii) 4L
(iii) 2 L
(iv) L
Answer (ii) 4L
Explanation:
The electric fields due to the two charges can cancel only outside the charges on the side of the smaller charge.
Using
$\frac{8}{x^2}$
$=$
$\frac{2}{(x-L)^2}$
Solving,
$x=2L$
and
$x=4L$
The valid position lies beyond the charge −2q-2q−2q.
Hence, the net electric field is zero at
$x=4L$
Q. Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be- (Sample Paper 2022)
(i) 5F
(ii) F
(iii) F/2
(iv) F/5
Answer (i) 5F
Explanation:
According to Coulomb’s law in a medium,
$F_m=\frac{F_v}{K}$
where
$K$
is the dielectric constant.
Given,
$K=5$
Therefore,
$F_v=5F_m$
Hence, the force in vacuum is
$5F$
Q. Which statement is true for Gauss law- (Sample Paper 2022)
(i) All the charges whether inside or outside the gaussian surface contribute to the electric flux.
(ii) Electric flux depends upon the geometry of the gaussian surface.
(iii) Gauss theorem can be applied to non-uniform electric field.
(iv) The electric field over the gaussian surface remains continuous and uniform at every point.
Answer (iii) Gauss theorem can be applied to non-uniform electric field.
Explanation:
The Gauss’s law is valid for any closed surface and for any charge distribution.
$\Phi=\frac{Q}{\varepsilon_0}$
Therefore, Gauss’s law can be applied even when the electric field is non-uniform.
Hence, option (iii) is correct.
Q. A cylinder of radius r and length l is placed in an uniform electric field parallel to the axis of the cylinder. The total flux for the surface of the cylinder is given by- (Sample Paper 2022)
(i) zero
(ii) π r2
(iii) E π r2
(iv) 2 Eπ r2
Answer (i) Zero
Explanation:
This is the same situation discussed in textbook for a closed cylindrical surface in a uniform electric field.
The flux entering one circular face is equal in magnitude and opposite in sign to the flux leaving the other face.
The flux through the curved surface is zero because the electric field is parallel to it.
Therefore,
$\Phi=0$
Q. Two parallel large thin metal sheets have equal surface densities 26.4 x 10-12 C/m2 of opposite signs. The electric field between these sheets is- (Sample Paper 2022)
(i) 1.5N/C
(ii) 1.5 x 10-16 N/C
(iii) 3 x 10-10N/C
(iv) 3N/C
Answer (iv) 3N/C
Explanation:
The electric field due to one infinite plane sheet is
$E=\frac{\sigma}{2\varepsilon_0}$
Between two oppositely charged sheets, the fields add.
Therefore,
$E=\frac{\sigma}{\varepsilon_0}$
Substituting,
$E=\frac{26.4\times10^{-12}}{8.854\times10^{-12}}$
$E\approx3,N/C$
Hence, the electric field is 3 N/C.
Q. Consider an uncharged conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then, (Sample Paper 2022)
(i) negative and uniformly distributed over the surface of sphere
(ii) positive and uniformly distributed over the surface of sphere
(iii) negative and appears at a point the surface of sphere closest to point charge.
(iv) Zero
Answer (iv) Zero
Explanation:
The conducting sphere is initially uncharged.
The external positive charge only redistributes the charges inside the conductor.
No charge is transferred to or from the sphere.
According to the conservation of charge, the net charge on the sphere remains zero.
Hence, the sphere remains electrically neutral.
Q. Three Charges 2q, -q and -q lie at vertices of a triangle. The value of E and V at centroid of triangle will be- (Sample Paper 2022)
(i) E#0 and V#0
(ii) E=0 and V=0
(iii) E#0 and V=0
(iv) E=0 and V#0
Answer (iii) E#0 and V=0
Explanation:
The centroid is equidistant from all three charges.
Potential is a scalar quantity and is added algebraically.
Therefore,
$V=$
$\frac{k(2q-q-q)}{r}$
$=0$
Electric field is a vector quantity.
The electric field vectors due to the three charges do not cancel each other.
Hence,
$E\neq0$
and
$V=0$
Therefore, the correct option is (iii).
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