1. Ten electrons are equally spaced and fixed around a circle of radius R. Relative to V = 0 at infinity, the electrostatic potential V and the electric field E at the centre C are
Correct Answer: (a) \(V = \frac{-10e}{4\pi\epsilon_0 R}\) and \(E = 0\)
The potential at the center is the sum of potentials due to all electrons: \(V = 10 \times \frac{-e}{4\pi\epsilon_0 R}\).
The electric field at the center is zero because the fields from opposite electrons cancel each other out due to symmetry.
2. Two point charges +q and -q are placed at a distance d apart. The electric potential at a point on the perpendicular bisector of the line joining the charges at distance r from the midpoint is:
Correct Answer: (c) Zero
The potential at any point due to a charge configuration is the algebraic sum of potentials due to individual charges.
For this dipole configuration on the perpendicular bisector, the distances to both charges are equal, so the potentials
\(+q\) and \(-q\) cancel each other out, resulting in zero net potential.
3. Consider the points lying on a straight line joining two fixed opposite charges. Between the charges there is
Correct Answer: (b) Only one point where electric field is zero
Between two opposite charges, there is exactly one point where the electric fields from both charges cancel each other out (E=0).
The potential is never zero between opposite charges as they have opposite signs and their contributions add up rather than cancel out.
4. A charged particle of mass m is held stationary in space by placing it in an electric field of strength E directed vertically downwards. The charge on the particle is
Correct Answer: (d) \(-\frac{mg}{E}\)
For the particle to be stationary, the electric force must balance the gravitational force: \(qE = mg\).
Since E is directed downward, the charge must be negative to produce an upward force.
Therefore, \(q = -\frac{mg}{E}\).
5. A hollow metal sphere of radius R is charged to a potential V. The electric field at a distance r from the center (r > R) is:
Correct Answer: (b) \(\frac{VR}{r^2}\)
For a charged sphere, outside the sphere (r > R), the potential is \(V = \frac{kQ}{r}\) and the electric field is \(E = \frac{kQ}{r^2}\).
Combining these gives \(E = \frac{VR}{r^2}\), since at the surface \(V = \frac{kQ}{R}\).
6. The dimension of (1/2)ε₀E² (ε₀: permittivity of free space; E: electric field) is
Correct Answer: (b) [ML⁻¹T⁻²]
The expression (1/2)ε₀E² represents energy density.
Dimensions of ε₀: [M⁻¹L⁻³T⁴A²],
Dimensions of E: [MLT⁻³A⁻¹],
So E² has dimensions [M²L²T⁻⁶A⁻²],
Thus ε₀E² has dimensions [ML⁻¹T⁻²], which is the same as pressure or energy density.
7. Potential at a point x-distance from the centre inside the conducting sphere of radius R and charged with charge Q is
Correct Answer: (a) \(\frac{1}{4\pi\epsilon_0} \frac{Q}{R}\)
Inside a conducting sphere, the potential is constant and equal to the potential at the surface, which is \(\frac{1}{4\pi\epsilon_0} \frac{Q}{R}\).
This is because the electric field inside a conductor is zero in electrostatic equilibrium.
8. A parallel plate capacitor with air between the plates has capacitance C. If the distance between the plates is doubled and the space between them is filled with a material of dielectric constant 4, the new capacitance will be:
Correct Answer: (b) 2C
Original capacitance \(C = \frac{\epsilon_0 A}{d}\).
New distance = 2d, dielectric constant K = 4.
New capacitance \(C' = \frac{K\epsilon_0 A}{2d} = \frac{4\epsilon_0 A}{2d} = 2 \times \frac{\epsilon_0 A}{d} = 2C\).
9. The work done in moving a charge Q from one point to another on an equipotential surface is:
Correct Answer: (b) Zero
On an equipotential surface, the potential difference between any two points is zero.
Work done \(W = Q\Delta V = Q \times 0 = 0\).
10. When a proton is accelerated through 1V, then its kinetic energy will be
Correct Answer: (c) 1 eV
The kinetic energy gained by a charge q accelerated through potential difference V is \(K = qV\).
For a proton (q = +e) through 1V, \(K = 1 \text{ eV}\).
11. An electron enters between two horizontal plates separated by 2mm and having a potential difference of 1000V. The force on electron is
Correct Answer: (a) \(8 \times 10^{-14} N\)
Electric field \(E = \frac{V}{d} = \frac{1000}{2 \times 10^{-3}} = 5 \times 10^5 \text{ V/m}\).
Force \(F = eE = 1.6 \times 10^{-19} \times 5 \times 10^5 = 8 \times 10^{-14} \text{ N}\).
12. The electric flux through a closed surface enclosing a charge q is:
Correct Answer: (a) \(\frac{q}{\epsilon_0}\)
This is Gauss's law: The total electric flux through any closed surface is equal to \(\frac{1}{\epsilon_0}\) times the net charge enclosed by the surface.
13. The number of electrons to be put on a spherical conductor of radius 0.1m to produce an electric field of 0.036 N/C just above its surface is
Correct Answer: (b) \(2.5 \times 10^6\)
Electric field at surface \(E = \frac{kQ}{R^2}\).
\(Q = \frac{ER^2}{k} = \frac{0.036 \times (0.1)^2}{9 \times 10^9} = 4 \times 10^{-14} \text{ C}\).
Number of electrons \(n = \frac{Q}{e} = \frac{4 \times 10^{-14}}{1.6 \times 10^{-19}} = 2.5 \times 10^5\).
14. Two positive point charges of 12μC and 8μC are 10cm apart. The work done in bringing them 4cm closer is
Correct Answer: (c) 13 J
Initial potential energy \(U_i = \frac{kq_1q_2}{r_1} = \frac{9 \times 10^9 \times 12 \times 8 \times 10^{-12}}{0.10} = 8.64 \text{ J}\).
Final distance = 6cm, \(U_f = \frac{9 \times 10^9 \times 12 \times 8 \times 10^{-12}}{0.06} = 14.4 \text{ J}\).
Work done \(W = U_f - U_i = 5.76 \text{ J}\) (closest to 5.8 J).
15. The capacitance of a parallel plate capacitor increases when:
Correct Answer: (c) A dielectric is inserted between plates
Capacitance \(C = \frac{K\epsilon_0 A}{d}\). It increases with:
(1) Larger plate area (A),
(2) Smaller separation (d),
(3) Higher dielectric constant (K).
Inserting a dielectric (K > 1) increases capacitance.
16. In an hydrogen atom, the electron revolves around the nucleus in an orbit of radius 0.53Å. Then the electrical potential produced by the nucleus at the position of the electron is
Correct Answer: (c) 27.2 V
Potential \(V = \frac{ke}{r} = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{0.53 \times 10^{-10}} = 27.2 \text{ V}\).
The sign is positive because it's the potential due to the proton's positive charge.
17. A proton is accelerated through 50,000 V. Its energy will increase by
Correct Answer: (b) \(8 \times 10^{-15} J\)
Energy gained \(E = qV = 50,000 \text{ eV} = 50,000 \times 1.6 \times 10^{-19} \text{ J} = 8 \times 10^{-15} \text{ J}\).
18. The electric potential V at any point O (x, y, z all in metres) in space is given by V = 4x². The electric field at the point (1m, 0, 2m) in V/m is
Correct Answer: (a) 8 along negative x-axis
Electric field \(E = -\frac{dV}{dx} = -\frac{d}{dx}(4x^2) = -8x\).
At (1,0,2), \(E = -8(1) = -8 \text{ V/m}\), which is 8 V/m along negative x-axis.
19. A charge Q is divided into two parts q and (Q-q). The maximum Coulomb repulsion between them occurs when:
Correct Answer: (b) q = Q/2
Force \(F = k\frac{q(Q-q)}{r^2}\). For maximum F, differentiate w.r.t q and set to zero:
\(\frac{dF}{dq} = k\frac{(1)(Q-q) + q(-1)}{r^2} = 0 \Rightarrow Q - 2q = 0 \Rightarrow q = Q/2\).
20. An electron of mass m and charge e is accelerated from rest through a potential difference V in vacuum. The final speed of the electron will be
Correct Answer: (b) \(\sqrt{\frac{2eV}{m}}\)
Kinetic energy gained = Potential energy lost: \(\frac{1}{2}mv^2 = eV\).
Solving for v: \(v = \sqrt{\frac{2eV}{m}}\).
21. The electric field intensity at a point due to an infinite plane sheet of charge with surface charge density σ is:
Correct Answer: (b) \(\frac{\sigma}{2\epsilon_0}\)
For an infinite plane sheet of charge, the electric field is constant and given by \(E = \frac{\sigma}{2\epsilon_0}\),
directed normally away from the sheet (for positive σ).
22. The radius of a soap bubble whose potential is 16V is doubled. The new potential of the bubble will be
Correct Answer: (c) 8V
Potential of a charged sphere \(V = \frac{kQ}{R}\). If radius doubles while charge remains same,
new potential \(V' = \frac{kQ}{2R} = \frac{V}{2} = \frac{16}{2} = 8 \text{ V}\).
23. If the electric flux entering and leaving a closed surface are ϕ₁ and ϕ₂ respectively, the charge enclosed in the surface is:
Correct Answer: (d) \(\frac{\phi_2 - \phi_1}{\epsilon_0}\)
Net flux \(\phi = \phi_2 - \phi_1\). By Gauss's law: \(\phi = \frac{Q_{enc}}{\epsilon_0}\).
Therefore, \(Q_{enc} = \epsilon_0 (\phi_2 - \phi_1)\).
24. Two charges +q and -q are placed at points A and B respectively which are a distance 2d apart. The electric field at the midpoint between A and B is:
Correct Answer: (d) \(\frac{2q}{4\pi\epsilon_0 d^2}\) towards B
At midpoint, fields due to both charges add up because they're in the same direction (from +q to -q).
\(E = \frac{q}{4\pi\epsilon_0 d^2} + \frac{q}{4\pi\epsilon_0 d^2} = \frac{2q}{4\pi\epsilon_0 d^2}\) towards B.
25. Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is E₀. When the space is filled with dielectric, the electric field becomes E₀/2. The dielectric constant of the dielectric material
Correct Answer: (c) 2
With dielectric, electric field becomes \(E = \frac{E_0}{K}\).
Given \(E = \frac{E_0}{2}\), so \(K = 2\).
26. Point charges +q, -q, and +q are kept on the x-axis at points x = -a, x = 0, and x = +a respectively, then
Correct Answer: (c) All the charges are in unstable equilibrium
The middle charge (-q) is in unstable equilibrium as any displacement would cause it to move towards one of the +q charges.
The outer +q charges are also in unstable equilibrium as any displacement would break the symmetry.
27. A particle A has charge +q and a particle B has charge +4q with each of them having the same mass m. When allowed to fall from rest through the same electric potential difference, the ratio of their speed vA/vB will be
Correct Answer: (b) 1/2
Kinetic energy gained = qV, so \(\frac{1}{2}mv_A^2 = qV\) and \(\frac{1}{2}mv_B^2 = 4qV\).
Taking ratio: \(\frac{v_A^2}{v_B^2} = \frac{1}{4} \Rightarrow \frac{v_A}{v_B} = \frac{1}{2}\).
28. The electric potential V is given as a function of distance x (metre) by V = (5x² + 10x - 9) V. Value of electric field at x = 1m is
Correct Answer: (a) -20 V/m
Electric field \(E = -\frac{dV}{dx} = -\frac{d}{dx}(5x^2 + 10x - 9) = -(10x + 10)\).
At x = 1m, \(E = -(10(1) + 10) = -20 \text{ V/m}\).
29. Equal charges q are placed at the vertices A and B of an equilateral triangle ABC of side a. The magnitude of electric field at the point C is
Correct Answer: (c) \(\frac{\sqrt{3}q}{4\pi\epsilon_0 a^2}\)
Each charge produces field \(E = \frac{q}{4\pi\epsilon_0 a^2}\) at C.
The angle between fields is 60°. Resultant field \(E_{net} = \sqrt{E^2 + E^2 + 2E^2 \cos 60°} = \sqrt{3}E = \frac{\sqrt{3}q}{4\pi\epsilon_0 a^2}\).
30. A charge Q is placed at the center of a cube. The flux through one face of the cube is:
Correct Answer: (b) \(\frac{Q}{6\epsilon_0}\)
Total flux through cube = \(\frac{Q}{\epsilon_0}\).
Since cube has 6 faces and by symmetry, flux through each face is equal,
flux through one face = \(\frac{Q}{6\epsilon_0}\).
31. Two insulated charged conducting spheres of radii R₁ and R₂ respectively and having an equal charge of Q are connected by a copper wire and then they are separated. Then
Correct Answer: (c) Surface charge density on the R₂ sphere will be greater than that on the R₁ sphere
When connected, potentials equalize: \(\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \Rightarrow \frac{Q_1}{Q_2} = \frac{R_1}{R_2}\).
Surface charge density \(\sigma = \frac{Q}{4\pi R^2}\), so \(\frac{\sigma_1}{\sigma_2} = \frac{Q_1/R_1^2}{Q_2/R_2^2} = \frac{R_2}{R_1}\).
Thus, smaller sphere (R₂) has higher charge density.
32. The energy stored in a parallel plate capacitor when a dielectric slab is inserted between the plates (without battery connected):
Correct Answer: (b) Decreases
With battery disconnected, charge Q remains constant.
Energy \(U = \frac{Q^2}{2C}\) and \(C\) increases with dielectric (\(C' = KC\)).
Thus \(U' = \frac{Q^2}{2KC} = \frac{U}{K}\), which is less than U since K > 1.
33. A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance d is
Correct Answer: (c) \(qEd\)
Work done by electric field = Change in kinetic energy.
\(W = Fd = qEd = \Delta K\).
Since initial KE = 0, final KE = qEd.
34. The capacitance of a spherical conductor of radius R is:
Correct Answer: (a) \(4\pi\epsilon_0 R\)
Capacitance of an isolated spherical conductor is \(C = \frac{Q}{V} = \frac{Q}{kQ/R} = 4\pi\epsilon_0 R\),
since \(k = \frac{1}{4\pi\epsilon_0}\).
35. A sphere of radius R has potential of V₀, then energy density near its surface will be
Correct Answer: (b) \(\frac{\epsilon_0 V_0^2}{2R^2}\)
Energy density \(u = \frac{1}{2}\epsilon_0 E^2\).
At surface, \(E = \frac{V_0}{R}\), so \(u = \frac{1}{2}\epsilon_0 \left(\frac{V_0}{R}\right)^2 = \frac{\epsilon_0 V_0^2}{2R^2}\).
36. Two unlike charges of magnitude q are separated by a distance d. The potential at a point midway between them is
Correct Answer: (a) Zero
Potential is scalar: \(V = V_+ + V_- = \frac{kq}{d/2} + \frac{k(-q)}{d/2} = \frac{2kq}{d} - \frac{2kq}{d} = 0\).
37. The electric field required to keep a water droplet of mass m and charge q stationary in the earth's gravitational field is:
Correct Answer: (a) \(\frac{mg}{q}\) upward
For equilibrium: \(qE = mg \Rightarrow E = \frac{mg}{q}\).
Direction must be upward to counteract gravity (assuming q is positive).
38. A capacitor of capacitance C is charged to potential V. The energy stored in it is U. If the potential is doubled, the new energy stored will be:
Correct Answer: (c) 4U
Energy stored \(U = \frac{1}{2}CV^2\).
If V becomes 2V, new energy \(U' = \frac{1}{2}C(2V)^2 = 4 \times \frac{1}{2}CV^2 = 4U\).
39. The equivalent capacitance between points A and B in the given network of capacitors (each of capacitance C) is:
Correct Answer: (b) 2C
The network consists of three capacitors: two in series (\(\frac{C}{2}\)) connected in parallel with the third (C).
Total capacitance = \(\frac{C}{2} + C + \frac{C}{2} = 2C\).
40. Two metal spheres of radii r₁ and r₂ are charged to the same potential. The ratio of charges on the spheres is:
Correct Answer: (b) \(\frac{r_1}{r_2}\)
Potential \(V = \frac{kQ}{r}\). For same V: \(\frac{Q_1}{r_1} = \frac{Q_2}{r_2} \Rightarrow \frac{Q_1}{Q_2} = \frac{r_1}{r_2}\).