JEE Physics Electric Charges and Fields MCQs Set 13

Answer: (d) \( \frac{q_1q_2}{56\sqrt{14}\,\pi\varepsilon_0}(\hat j-2\hat i-3\hat k) \)

Explanation:
1. The vector from q₁ to q₂ is \( \vec r=(0-2)\hat i+(0+1)\hat j+(0-3)\hat k=-2\hat i+\hat j-3\hat k \) and \( |\vec r|=\sqrt{14} \).
2. \( \vec F=\frac{q_1q_2}{4\pi\varepsilon_0}\frac{\vec r}{r^3}=\frac{q_1q_2}{4\pi\varepsilon_0\cdot14\sqrt{14}}(\hat j-2\hat i-3\hat k) \).
3. Since 4 × 14 = 56, this is option (d).

Teacher's Note:
1. Force on q₂ is along the vector from q₁ to q₂ for like charges.
2. Use F = kq₁q₂ r⃗ / r³ so that the denominator has r³ = 14√14.

Answer: (a) 8 : 1

Explanation:
1. Initial force ∝ 40 × 20 = 800.
2. After touching, each sphere carries (40 − 20)/2 = 10 μC, so the final force ∝ 10 × 10 = 100.
3. Ratio = 800 : 100 = 8 : 1.

Teacher's Note:
1. Identical conductors share the total charge equally on contact.
2. Compare the products of the charges at the same distance.

Answer: (d) \( 2\pi\sqrt{\frac{\ell}{\sqrt{g^2+(qE/m)^2}}} \)

Explanation:
1. The bob feels gravity mg downward and an electric force qE horizontally, at right angles to each other.
2. The effective acceleration is \( g_{eff}=\sqrt{g^2+(qE/m)^2} \).
3. Hence \( T=2\pi\sqrt{\ell/g_{eff}} \).

Teacher's Note:
1. Perpendicular forces combine by Pythagoras.
2. Replace g by the effective g in the pendulum formula.

Answer: (c) \( \frac{1}{4\pi\varepsilon_0}\frac{2q}{3\sqrt3R^2} \)

Explanation:
1. On the axis \( E=\frac{kqx}{(R^2+x^2)^{3/2}} \); setting dE/dx = 0 gives x = R/√2.
2. Then \( E_{max}=\frac{kq\,R/\sqrt2}{(3R^2/2)^{3/2}}=\frac{2kq}{3\sqrt3R^2} \).

Teacher's Note:
1. Differentiate E with respect to the axial distance x.
2. The maximum occurs at x = R/√2.

Answer: (c) both at P and at Q

Explanation:
1. At P (axial point) the field is along the dipole moment, parallel to the axis.
2. At Q (equatorial point) the field is antiparallel to the dipole moment, again parallel to the axis.
3. So the field is parallel to the axis at both points.

Teacher's Note:
1. The field on the axial line and on the equatorial line is always along the axis direction.
2. Only the sense differs: same direction on axial, opposite on equatorial.

Answer: (c) 2.0 × 10⁻³ N·m

Explanation:
1. p = qd = 1.0 × 10⁻⁶ × 2.0 × 10⁻² = 2.0 × 10⁻⁸ C·m.
2. τ_max = pE = 2.0 × 10⁻⁸ × 1.0 × 10⁵ = 2.0 × 10⁻³ N·m.

Teacher's Note:
1. Maximum torque occurs when the dipole is perpendicular to the field.
2. Convert 2.0 cm to metres.

Answer: (c) neither the electric field nor the electric potential is zero

Explanation:
1. On the axial line \( E=\frac{2kp}{r^3} \) along the axis and \( V=\pm\frac{kp}{r^2} \).
2. Both are non-zero at a finite distance.

Teacher's Note:
1. Use the axial-point formulas for E and V.
2. V vanishes only on the equatorial line.

Answer: (c) Electric lines of force are perpendicular to an equipotential surface

Explanation:
1. Along an equipotential surface no work is done, so the field has no component along it.
2. Hence lines of force cross equipotential surfaces at right angles.
3. Electrostatic lines are open (start on positive, end on negative charges), and a released charge follows a line only when the line is straight.

Teacher's Note:
1. Field = −gradient of potential, which is normal to equipotentials.
2. Lines of force are not generally particle trajectories.

Answer: (c) The interior remains neutral and the surface gets a non-uniform charge distribution

Explanation:
1. The sphere as a whole stays neutral (no contact).
2. Charges separate on the surface: negative near q, positive on the far side, so the distribution is non-uniform.
3. The field inside a conductor is zero, so the interior stays neutral.

Teacher's Note:
1. Electrostatic induction moves charge on the surface only.
2. Net charge on the sphere remains zero.

Answer: (c) 50 V/m

Explanation:
1. For the non-conducting sheet \( E=\frac{\sigma}{2\varepsilon_0} \), where σ = q/A.
2. On a conducting plate the charge spreads equally on both faces, each face having σ/2, and the field outside is \( \frac{\sigma/2}{\varepsilon_0}=\frac{\sigma}{2\varepsilon_0} \).
3. So the field stays 50 V/m.

Teacher's Note:
1. A conductor's charge splits equally on its two faces.
2. The field just outside a conductor is (surface density)/ε₀.

Answer: (b) The lines bend to meet the sphere surface at right angles and no lines enter its interior

Explanation:
1. The conductor's induced charges cancel the field inside it, so no lines of force enter the interior.
2. Field lines must be normal to a conducting surface, so they bend to end (and restart) perpendicular to the sphere's surface.

Teacher's Note:
1. Field inside a conductor is zero.
2. Field at the surface of a conductor is always normal to it.

Answer: (a) remain in the car

Explanation:
1. The metal body of the car acts as a conducting shell (Faraday cage).
2. The field inside the shell is zero, and the charge flows around the outside to the ground.

Teacher's Note:
1. A closed conductor shields its interior.
2. Trees and poles are poor choices because the discharge passes through them.

Answer: (d) none of these

Explanation:
1. Between A and B both fields point the same way, so there is no balance.
2. Outside AB, the nearer charge always dominates because the magnitudes are equal, so the fields never cancel.
3. On the perpendicular bisector the net field is parallel to AB and non-zero.
4. So there is no such point.

Teacher's Note:
1. Check the three regions: between, outside and the bisector.
2. Equal and opposite charges give zero field nowhere at a finite distance.

Answer: (c) 3/4

Explanation:
1. Charge in the cube = area under the density graph × 1 m² = ρ₀(½·¼ + ½ + ½·¼) = 0.75ρ₀.
2. Flux = Q/ε₀ = 0.75ρ₀/ε₀ = 0.75 since ρ₀ = ε₀ numerically.
3. Hence the flux is 3/4.

Teacher's Note:
1. Total charge is the area under the density curve per unit area.
2. Gauss's law: flux = enclosed charge / ε₀.

Answer: (c) along OC

Explanation:
1. The upper (positive) arc pushes the field at O down and to the right.
2. The lower (negative) arc pulls the field at O down and to the left.
3. The horizontal components cancel by symmetry and the vertical components add downward.
4. So the field points along OC.

Teacher's Note:
1. Split the ring into the two halves and add the vectors.
2. Symmetry cancels the components along OB and OD.

Answer: (a) a is positive, b is negative

Explanation:
1. If both were of the same sign the field would reverse direction at the middle and cross zero.
2. The field is directed from a towards b throughout, which happens when a is positive (field points away from it) and b is negative (field points towards it).

Teacher's Note:
1. A field that never changes sign between the charges means unlike charges.
2. Field points away from a positive charge and towards a negative one.

Answer: (b) \( \frac{-2kP_1P_2\cos\theta}{r^3} \)

Explanation:
1. The field of P₂ at P₁'s position has a radial component \( \frac{2kP_2\cos\theta}{r^3} \) and a transverse component \( \frac{kP_2\sin\theta}{r^3} \).
2. P₁ points along the radial direction, so \( U=-\vec P_1\cdot\vec E=-\frac{2kP_1P_2\cos\theta}{r^3} \).

Teacher's Note:
1. U = −p·E; only the radial component of the field matters here.
2. Use the dipole field in polar form.

Answer: (a) increase

Explanation:
1. The electric force qE acts upward, opposing gravity.
2. Effective gravity is \( g'=g-\frac{qE}{m}<g \).
3. Since \( T=2\pi\sqrt{\ell/g'} \), the period increases.

Teacher's Note:
1. Upward electric force reduces effective g.
2. Smaller g gives a longer period.

Answer: (d) 1/x⁴

Explanation:
1. The net charge is zero and the dipole moment is also zero, so the leading term is the quadrupole term.
2. Directly: \( E_x=\frac{2kqx}{(x^2+d^2)^{3/2}}-\frac{2kq}{x^2}\approx-\frac{3kqd^2}{x^4} \).
3. So E varies as 1/x⁴.

Teacher's Note:
1. Expand (x² + d²)^(−3/2) for x ≫ d.
2. The 1/x² and 1/x³ terms cancel.

Answer: (c) \( E=\frac{Ze}{4\pi\varepsilon_0}\left(\frac1{r^2}-\frac r{R^3}\right)\ (r<R),\ \ 0\ (r>R) \)

Explanation:
1. Outside (r > R) the atom is neutral, so E = 0.
2. Inside, the nucleus gives \( \frac{Ze}{4\pi\varepsilon_0r^2} \) outward, and the negative cloud enclosed within r is \( -Ze\frac{r^3}{R^3} \), giving \( -\frac{Ze\,r}{4\pi\varepsilon_0R^3} \).
3. So \( E=\frac{Ze}{4\pi\varepsilon_0}\left(\frac1{r^2}-\frac r{R^3}\right) \).

Teacher's Note:
1. Use Gauss's law with the enclosed charge at each radius.
2. The negative charge inside radius r is proportional to r³.

Answer: (d) 3F/8

Explanation:
1. Let each sphere have charge q, so F = kq²/d².
2. Touching the third sphere to the first: both have q/2. Then touching it to the second: (q/2 + q)/2 = 3q/4 each.
3. New force = k(q/2)(3q/4)/d² = 3F/8.

Teacher's Note:
1. Identical conductors share their total charge equally on contact.
2. Do the two contacts one after the other.

Answer: (b) the electric field everywhere on the θ = π/2 plane is normal to the plane

Explanation:
1. At θ = π/2, \( E_r=-\frac{\partial V}{\partial r}=\frac{2p\cos\theta}{4\pi\varepsilon_0r^3}=0 \) but \( E_\theta=-\frac1r\frac{\partial V}{\partial\theta}=\frac{p\sin\theta}{4\pi\varepsilon_0r^3}\ne0 \).
2. The θ-direction on that plane is perpendicular to the plane, so the field is normal to it (and non-zero).
3. On the θ = 0 line the field is 2kp/r³, not zero.

Teacher's Note:
1. A zero potential does not mean zero field.
2. Equipotentials are perpendicular to the field, so the field on the V = 0 plane is normal to it.

Answer: (a) E increases linearly with r up to R, reaches its maximum value at r = R and then decreases as 1/r²

Explanation:
1. Inside, Gauss's law gives \( E=\frac{\rho r}{3\varepsilon_0} \), proportional to r.
2. Outside, \( E=\frac{kQ}{r^2} \).
3. The field is largest at r = R, where the two expressions meet.

Teacher's Note:
1. Enclosed charge inside radius r ∝ r³.
2. The field is continuous at the surface of a uniformly charged (non-conducting) sphere.

Answer: (d) is only that due to +q

Explanation:
1. The conducting shell screens the cavity from the outside, so +Q has no effect inside.
2. The field in the cavity is due to +q and the induced charge on the inner surface only.

Teacher's Note:
1. Electrostatic shielding works from outside to inside.
2. Induced charges adjust to cancel external fields in the cavity.

Answer: (b) \( \left[\frac{\sqrt3}{12\pi\varepsilon_0}\frac{q^2}{\kappa}\right]^{1/3} \)

Explanation:
1. Let the distance from the centre be r; the side of the triangle is √3 r.
2. The repulsion from the other two charges, resolved towards the line to the centre, gives \( F_{rep}=2\cdot\frac{kq^2}{3r^2}\cos30^\circ=\frac{kq^2}{\sqrt3\,r^2} \).
3. Equilibrium: \( \kappa r=\frac{q^2}{4\sqrt3\pi\varepsilon_0r^2} \), so \( r^3=\frac{q^2}{4\sqrt3\pi\varepsilon_0\kappa}=\frac{\sqrt3}{12\pi\varepsilon_0}\frac{q^2}{\kappa} \).

Teacher's Note:
1. Resolve the two repulsions along the radius.
2. Equate the net outward force to κr.

Answer: (c) L/3

Explanation:
1. Range ∝ 1/g_eff. For the neutral particle g_eff = g.
2. For charge q: range L/2 means g_eff = 2g, so the electric force adds an acceleration g (downward).
3. For charge 2q the added acceleration is 2g, so g_eff = 3g and the range is L/3.

Teacher's Note:
1. Range R = u² sin 2θ / g_eff.
2. The electric acceleration is proportional to the charge.

Answer: (d) parallel

Explanation:
1. A uniform field has the same magnitude and direction everywhere.
2. Lines of force then are equally spaced parallel straight lines.

Teacher's Note:
1. The density of lines measures the field strength.
2. Constant direction and strength means parallel, equidistant lines.