JEE Physics Electric Charges and Fields MCQs Set 25

Answer: (b) \( \frac{\sqrt2\,Q^2}{8\pi\epsilon_0R^2} \)

Explanation:
1. Each fixed charge is at distance \( \sqrt2R \), so each exerts \( \frac{kQ^2}{2R^2} \) at 45° to the x-axis.
2. The y-components cancel; the x-components add: \( 2\cdot\frac{kQ^2}{2R^2}\cdot\frac1{\sqrt2} = \frac{kQ^2}{\sqrt2R^2} = \frac{\sqrt2Q^2}{8\pi\epsilon_0R^2} \).

Teacher's Note:
1. Resolve each force into components and use symmetry to cancel one pair.
2. Distance from each fixed charge is the hypotenuse of a right triangle with both legs R.

Answer: (a) \( \frac{\lambda_0L}{2} \)

Explanation:
1. \( Q = \int_0^L\frac{\lambda_0x}{L}dx = \frac{\lambda_0}{L}\cdot\frac{L^2}{2} = \frac{\lambda_0L}{2} \).

Teacher's Note:
1. Integrate the density over the length when it is not uniform.
2. The average density is half the maximum value, so the charge is half of λ₀L.

Answer: (c) Newton's law of gravitation

Explanation:
1. Both forces vary as the inverse square of the distance and act along the line joining the two bodies, with the product of charges (masses) in the numerator.

Teacher's Note:
1. The analogy is mathematical: k q₁q₂/r² against G m₁m₂/r².
2. Gravity is always attractive, while the electric force can attract or repel.

Answer: (c) \( \frac{R}{2\sqrt2} \)

Explanation:
1. With a = R/2, the field on the bisector is \( E = \frac{2kqx}{(x^2+a^2)^{3/2}} \).
2. Setting dE/dx = 0 gives \( x = \frac{a}{\sqrt2} = \frac{R}{2\sqrt2} \).

Teacher's Note:
1. Maximise the force expression by differentiating with respect to x.
2. The field is zero at the centre and at infinity, so it must peak somewhere in between.

Answer: (d) along the negative y-direction

Explanation:
1. Take A at the origin, B at \( (\tfrac{\sqrt3}2a, \tfrac a2) \), C at \( (\tfrac{\sqrt3}2a, -\tfrac a2) \).
2. B repels A along (−√3/2, −1/2); C attracts A along (√3/2, −1/2).
3. The x-components cancel and the y-components add, giving a force of magnitude F along −y.

Teacher's Note:
1. Draw the two forces separately and add them as vectors.
2. Like charges repel and unlike charges attract.

Answer: (b) The total charge of the universe is constant

Explanation:
1. Charge is conserved: only the algebraic sum of charges stays fixed.
2. Pair production and annihilation change the number of charged particles and the amounts of positive and negative charge separately, but not the net charge.

Teacher's Note:
1. Conservation of charge is about the net (algebraic) charge only.
2. Charged particles can be created or destroyed in pairs.

Answer: (b) The magnitudes of the electric forces on them are equal

Explanation:
1. F = eE for both particles, so the magnitudes are equal but the directions are opposite.
2. Accelerations are eE/m, which differ because the electron is much lighter.

Teacher's Note:
1. Forces are vectors, so opposite directions mean the forces are not equal.
2. Acceleration depends on mass as well as on force.

Answer: (a) \( \frac{2kQq_0}{r^2} \), directed from O towards E

Explanation:
1. Five equal charges +Q at the vertices would give zero net force at O.
2. Replacing the +Q at E by −Q is equivalent to adding a charge −2Q at E.
3. The force of −2Q on a positive q₀ is attractive, \( \frac{2kQq_0}{r^2} \), directed towards E.

Teacher's Note:
1. Use the superposition idea: symmetric arrangement gives zero, then correct the one changed charge.
2. The force is toward the negative charge for a positive test charge.

Answer: (d) \( \frac{Q}{2\epsilon_0L^2} \)

Explanation:
1. Radius R = L/π and λ = Q/L. Field at the centre of a semicircular arc: \( E = \frac{2k\lambda}{R} \).
2. \( E = \frac{2kQ\pi}{L^2} = \frac{Q}{2\epsilon_0L^2} \).

Teacher's Note:
1. Only components along the axis of symmetry survive; those perpendicular cancel.
2. Express R in terms of the given length L.

Answer: (b) \( \frac{3kQ}{2a^2} \)

Explanation:
1. The two adjacent charges (distance a, 120° between their fields) give a resultant \( \frac{kQ}{a^2} \) along the axis.
2. The opposite charge 2Q is at distance 2a: \( \frac{k(2Q)}{4a^2} = \frac{kQ}{2a^2} \), along the same direction.
3. Total \( = \frac{3kQ}{2a^2} \).

Teacher's Note:
1. Two equal vectors at 120° to each other have a resultant equal to either one.
2. The distance to the opposite vertex of a regular hexagon is twice the side.

Answer: (a) \( \frac{q}{\sqrt2\pi\epsilon_0r^2} \)

Explanation:
1. Each charge gives a field \( kq/r^2 \) directed away from a positive and towards a negative charge.
2. Adding components, the x-components cancel and the y-components add to 2√2 kq/r² along the y-axis.
3. \( E = 2\sqrt2\cdot\frac{q}{4\pi\epsilon_0r^2} = \frac{q}{\sqrt2\pi\epsilon_0r^2} \).

Teacher's Note:
1. Draw the eight field vectors before adding them.
2. A negative charge at angle θ acts like a positive charge at θ + 180°.

Answer: (c) \( \alpha \)-particles

Explanation:
1. Only charged particles experience a force in an electric field.
2. X-rays and γ-rays are photons and neutrons are neutral, so only the α-particle is deflected.

Teacher's Note:
1. Deflection needs a charge.
2. Photons carry no charge.

Answer: (c) \( \frac{5\sqrt2\lambda}{4\pi\epsilon_0r} \), directed slightly above the +x direction

Explanation:
1. A quarter arc of density λ gives a field \( \sqrt2k\lambda/r \) along its bisector.
2. Left quarter (+3λ): \( \frac{3k\lambda}{r}(1,-1) \). Right quarter (−4λ): field points towards it, \( \frac{4k\lambda}{r}(1,1) \).
3. Sum \( = \frac{k\lambda}{r}(7,1) \), magnitude \( 5\sqrt2\frac{k\lambda}{r} \), mostly along +x with a small +y component.

Teacher's Note:
1. Resolve each quarter's field into x and y components and add.
2. Direction of the field is away from positive and towards negative charge.

Answer: (d) E/25

Explanation:
1. OA = 5 m, OB = \( \sqrt{24^2+7^2} = 25 \) m.
2. Field \( \propto 1/r^2 \): \( E' = E\cdot\left(\frac{5}{25}\right)^2 = \frac{E}{25} \).

Teacher's Note:
1. Only distance matters for the magnitude of the field of a point charge.
2. Distance to B is 5 times that to A, so the field falls by 25.

Answer: (b) \( \frac{\sqrt3\lambda R}{\epsilon_0} \)

Explanation:
1. A chord at distance R/2 has length \( 2\sqrt{R^2-R^2/4} = \sqrt3R \) and carries charge \( \sqrt3\lambda R \).
2. Gauss's law: \( \phi = \frac{q}{\epsilon_0} = \frac{\sqrt3\lambda R}{\epsilon_0} \).

Teacher's Note:
1. Find the charge enclosed first, then apply Gauss's law.
2. Use Pythagoras to get the half-length of the chord.

Answer: (c) \( \sqrt3:1 \)

Explanation:
1. Flux \( = \lambda L_{inside}/\epsilon_0 \). Longest length inside is the space diagonal \( \sqrt3\,\ell \); shortest is the edge length ℓ.
2. Ratio \( = \sqrt3:1 \).

Teacher's Note:
1. Flux depends only on the enclosed charge, so only the length of wire inside the cube matters.
2. Maximum path inside a cube is its body diagonal.

Answer: (a) B and C only

Explanation:
1. Flat face area is \( \pi r^2\times10^{-4} \) m², so (A) is wrong (flux \( E_0\pi r^2\times10^{-4} \)).
2. The field is along the axis, so the curved surface flux is zero: (B) is correct.
3. Total flux \( = 2E_0\pi r^2\times10^{-4} \); \( q = \epsilon_0\phi = 8.85\times10^{-12}\times6.28\times10^{-4}E_0r^2 \approx 5.5\times10^{-15}E_0r^2 \): (C) is correct.

Teacher's Note:
1. Convert cm² to m² carefully: 1 cm² = 10⁻⁴ m².
2. Use q = ε₀ × net flux.

Answer: (d) φ₁ = φ₂

Explanation:
1. Flux depends on the solid angle subtended at the charge; the field falls as 1/r² while the area grows as r², so the product is constant.
2. Both surfaces cut the same cone, so the flux is equal.

Teacher's Note:
1. Flux through a surface equals (q/ε₀) × (solid angle / 4π).
2. Distance does not matter for surfaces within the same cone.

Answer: (d) \( \frac{q}{6\epsilon_0} \)

Explanation:
1. Each vertex charge sends \( \frac{q}{8\epsilon_0} \) through the cube's surface; total \( = \frac{q}{\epsilon_0} \).
2. By symmetry each of the 6 faces gets \( \frac{q}{6\epsilon_0} \).

Teacher's Note:
1. A vertex charge has one eighth of its flux passing through the cube.
2. The eight charges are placed symmetrically, so all faces receive equal flux.

Answer: (a) E on the left side comes from all charges, while q on the right side comes from q₂ and q₄ only

Explanation:
1. The electric field at the surface is due to all charges, inside and outside.
2. Only the enclosed charges q₂ and q₄ contribute to the net flux.

Teacher's Note:
1. Outside charges produce flux that enters and leaves, giving no net contribution.
2. E is the total field; q is only the enclosed charge.

Answer: (c) \( \frac{\sigma}{2\epsilon_0} \), directed outwards

Explanation:
1. Just outside the shell \( E = \sigma/\epsilon_0 \); just inside it is zero.
2. The field in the hole is the average of the two, \( \frac{\sigma}{2\epsilon_0} \), directed outwards for positive σ.

Teacher's Note:
1. The missing patch itself produces the discontinuity, so the hole field is the average.
2. Positive charge gives an outward field.

Answer: (a) I: attraction, II: repulsion

Explanation:
1. In case I, opposite charges are nearest (head-to-tail), so they attract.
2. In case II, like (positive) ends face each other, so they repel.

Teacher's Note:
1. Look at which ends of the dipoles are closest.
2. Head-to-tail alignment lowers energy.

Answer: (d) \( r^{-4} \)

Explanation:
1. Axial field \( \propto 1/r^3 \). Force \( = p\,\frac{dE}{dr} \propto 1/r^4 \).

Teacher's Note:
1. The force on a dipole in a non-uniform field is p dE/dr.
2. One extra power of r comes from the differentiation.

Answer: (b) \( \vec E_A = -2\vec E_B \)

Explanation:
1. Axial: \( E_A = \frac{2kp}{r^3} \) along p. Equatorial: \( E_B = \frac{kp}{r^3} \) opposite to p.
2. So \( \vec E_A = -2\vec E_B \).

Teacher's Note:
1. The equatorial field is antiparallel to the dipole moment.
2. Axial field is twice the equatorial field at equal distance.

Answer: (c) only a torque

Explanation:
1. The forces on the two charges are equal and opposite, so the net force is zero.
2. They act along different lines, so they produce a torque p × E.

Teacher's Note:
1. Zero net force does not mean zero torque.
2. Torque makes the dipole align with the field.

Answer: (a) a force towards the left

Explanation:
1. The −q at the left is in a stronger field than the +q at the right.
2. The force on −q (towards the left) is larger than the force on +q (towards the right), so there is a net force to the left.

Teacher's Note:
1. A dipole is pulled towards the region of stronger field.
2. Net force exists only in a non-uniform field.

Answer: (d) \( -\frac{E_0}{16} \)

Explanation:
1. Axial: \( E_0 = \frac{2kp}{r^3} \). Equatorial at 2r: \( \frac{kp}{8r^3} = \frac{E_0}{16} \).
2. The equatorial field is opposite to the axial direction, so it is −E₀/16.

Teacher's Note:
1. Equatorial field is half the axial field at the same distance and opposite in direction.
2. The field falls as the cube of the distance, giving a factor 8.

Answer: (d) q₁ > q₂ and m₁ < m₂

Explanation:
1. Mutual forces are equal and opposite, so tan θ = F/(mg) gives tan θ ∝ 1/m.
2. A larger angle means smaller mass, so m₁ < m₂; the charge values can be anything.

Teacher's Note:
1. The charge magnitudes drop out because the force is the same on both.
2. Lighter ball swings farther.