JEE Physics Electric Charges and Fields MCQs Set 04

Answer: (d) All of them

Explanation:
1. Charge is quantised: q = ne with e = 1.6 × 10⁻¹⁹ C and n a whole number.
2. 4.0/1.6 = 2.5, 6.0/1.6 = 3.75 and 10.0/1.6 = 6.25 are not whole numbers.
3. So none of the three values can occur.

Teacher's Note:
1. Every observed charge is a whole-number multiple of the electron charge.
2. Divide the given charge by 1.6 × 10⁻¹⁹ C and check whether you get an integer.

Answer: (d) zero

Explanation:
1. The sphere was neutral and no charge has been added to or removed from it.
2. The outside charge only separates the sphere's charges (induction): the near side becomes negative and the far side positive.
3. The two induced charges are equal and opposite, so the net charge is zero.

Teacher's Note:
1. Induction redistributes charge but does not change the total charge of an isolated body.
2. Do not confuse induced surface charge with net charge.

Answer: (a) \( \frac{3}{4}F \)

Explanation:
1. Initially F = kq²/r².
2. After transfer the charges are q/2 and q + q/2 = 3q/2.
3. New force F' = k(q/2)(3q/2)/r² = (3/4)kq²/r² = (3/4)F.

Teacher's Note:
1. The product of charges matters, not their sum.
2. Here the sum stays 2q but the product falls from q² to 3q²/4.

Answer: (c) \( r/\sqrt{K} \)

Explanation:
1. In air, F = kq₁q₂/r².
2. In the medium, F' = kq₁q₂/(K r'²).
3. Setting F' = F gives K r'² = r², so r' = r/√K.

Teacher's Note:
1. A dielectric reduces the force by the factor K, so the charges must be brought closer.
2. Equal forces need Kr'² = r².

Answer: (b) 12 cm from +9q

Explanation:
1. Q is in equilibrium where the forces from the two charges are equal: 9q/x² = q/(16 − x)².
2. Taking the square root: 3/x = 1/(16 − x), so 48 − 3x = x.
3. x = 12 cm from +9q (and 4 cm from +q).

Teacher's Note:
1. For equilibrium of the third charge, the nearer (smaller) charge must be closer.
2. Check: 9/144 = 1/16 = 0.0625, so the forces balance.

Answer: (a) \( F \)

Explanation:
1. Initially the product of charges has magnitude 7 × 5 = 35 (µC)².
2. After adding −2 µC to each, the charges are +5 µC and −7 µC.
3. The product is again 35 (µC)² with unlike signs, so the force is attractive and equal to F.

Teacher's Note:
1. Coulomb force depends only on the product q₁q₂ and the distance.
2. Here the two charges simply exchange their values.

Answer: (a) 0.2

Explanation:
1. F = kq²/r², so q² = Fr²/k = 0.144 × (0.05)²/(9 × 10⁹) = 4 × 10⁻¹⁴.
2. q = 2 × 10⁻⁷ C = 0.2 µC.

Teacher's Note:
1. Convert 5 cm to 0.05 m before substituting.
2. Square-root of 4 × 10⁻¹⁴ is 2 × 10⁻⁷.

Answer: (a) 4 N (attractive)

Explanation:
1. Initially 12 N = k(2 × 6)(10⁻¹²)/r², so r² = 9 × 10⁹ × 12 × 10⁻¹²/12 = 9 × 10⁻³ m².
2. New charges are −2 µC and +2 µC, which attract.
3. F' = 9 × 10⁹ × (2 × 2) × 10⁻¹²/(9 × 10⁻³) = 4 N, attractive.

Teacher's Note:
1. Compare with the original situation instead of finding r numerically: the product of charges falls from 12 to 4 (in µC²) so the force becomes one third of 12 N.
2. The sign of the product decides attraction or repulsion.

Answer: (a) \( 8.6 \times 10^{13}\ \text{C} \)

Explanation:
1. Set kq²/r² = GMm/r², so q² = GMm/k.
2. q² = 6.67 × 10⁻¹¹ × 10²⁵ × 10²³/(9 × 10⁹) = 7.41 × 10²⁷.
3. q = 8.6 × 10¹³ C.

Teacher's Note:
1. Both forces vary as 1/r², so the distance cancels.
2. Use k = 9 × 10⁹ N m² C⁻².

Answer: (a) \( F \)

Explanation:
1. Coulomb forces obey the principle of superposition.
2. The presence of other charges does not change the force between a given pair.
3. So the force of q₁ on q₂ remains F (only the net force on q₂ changes).

Teacher's Note:
1. Be careful to separate the force due to one charge from the net force on a charge.
2. Superposition means pairwise forces simply add.

Answer: (d) \( Q_1 = Q_2 = Q/2 \)

Explanation:
1. F = kQ₁Q₂/R² with Q₁ + Q₂ = Q fixed.
2. The product Q₁Q₂ = Q₁(Q − Q₁) is maximum when Q₁ = Q/2.
3. Hence Q₁ = Q₂ = Q/2.

Teacher's Note:
1. For a fixed sum, the product of two numbers is greatest when they are equal.
2. Check with numbers: (Q/2)² = 0.25Q² beats (2Q/3)(Q/3) = 0.22Q².

Answer: (b) \( -q \)

Explanation:
1. Force on q due to 4q: k(4q)(q)/l², directed away from 4q.
2. Force on q due to Q (distance l/2): kQq/(l/2)² = 4kQq/l².
3. For zero net force Q must attract q, so Q is negative, and 4|Q| = 4q gives Q = −q.

Teacher's Note:
1. Draw the three charges in a line and mark the direction of each force on q.
2. Zero net force needs equal magnitude and opposite directions.

Answer: (c) 8F towards the −q charge

Explanation:
1. Let the separation be d, so F = kq²/d².
2. The new sphere is at distance d/2 from each; each force has magnitude kq²/(d/2)² = 4F.
3. The force from +q is repulsive (towards −q) and the force from −q is attractive (also towards −q), so they add: 8F towards −q.

Teacher's Note:
1. Forces from a like charge and an unlike charge on the middle charge point the same way.
2. Halving the distance multiplies the force by 4.

Answer: (a) \( 180^\circ,\ \frac{1}{4\pi\varepsilon_0}\frac{Q^2}{(2L)^2} \)

Explanation:
1. With no gravity the only forces are the tension and the mutual repulsion, so the threads lie along one straight line: the angle is 180°.
2. The spheres are then 2L apart.
3. Tension equals the repulsion: T = Q²/(4πε₀(2L)²).

Teacher's Note:
1. Without weight to pull the spheres down, the repulsion is balanced by tension alone.
2. The separation is 2L, not L.

Answer: (b) \( \frac{\sqrt{3}\,r}{\sqrt{3}+1} \) from 9e

Explanation:
1. The field is zero between the like charges at distance x from 9e: 9/x² = 3/(r − x)².
2. So √3 (r − x) = x, giving x = √3 r/(1 + √3) from the 9e charge.
3. (This is closer to the 3e charge, which is the weaker one: distance from 3e is r/(1 + √3).)

Teacher's Note:
1. The zero-field point lies closer to the smaller charge.
2. Options (1) and (2) differ only in how the distance from 9e is written.

Answer: (b) \( 3.6 \times 10^{4} \)

Explanation:
1. At the midpoint (0.5 m from each) the two fields are opposite.
2. E = k(2 × 10⁻⁶ − 1 × 10⁻⁶)/(0.5)² = 9 × 10⁹ × 10⁻⁶/0.25.
3. E = 3.6 × 10⁴ N/C, directed away from the 2 µC charge, i.e. towards the 1 µC charge.

Teacher's Note:
1. Fields from two positive charges oppose each other between them, so subtract.
2. Distance is 0.5 m, not 1 m.

Answer: (c) \( ML^{4}T^{-3}A^{-1} \)

Explanation:
1. K = E x³.
2. [E] = force/charge = MLT⁻²/(AT) = MLT⁻³A⁻¹.
3. [K] = MLT⁻³A⁻¹ × L³ = ML⁴T⁻³A⁻¹.

Teacher's Note:
1. Dimensions of E are those of force per unit charge, with charge = A·T.
2. Multiply by L³ from x³.

Answer: (d) 32 V/m

Explanation:
1. For a semicircular arc the field at the centre is E = 2kλ/R with λ = Q/(πR).
2. E = 2kQ/(πR²) = 2 × 9 × 10⁹ × 1.4 × 10⁻⁹/(π × 0.25).
3. E = 25.2/0.785 ≈ 32 V/m.

Teacher's Note:
1. Only the components along the axis of symmetry add; those perpendicular cancel.
2. The standard result is E = 2kλ/R for a semicircle.

Answer: (b) 10 V/m downward

Explanation:
1. The electric force must be upward to balance the weight: |q|E = mg.
2. E = mg/|q| = 10⁻⁶ × 10/10⁻⁶ = 10 V/m.
3. The charge is negative, so the force is opposite to E; an upward force needs E downward.

Teacher's Note:
1. Force on a negative charge is opposite to the field.
2. Compute magnitude first, then fix direction.

Answer: (a) \( \tan^{-1}(0.1) \)

Explanation:
1. Electric force F = kq²/r² = 9 × 10⁹ × 10⁻¹⁸/(3 × 10⁻³)² = 10⁻³ N.
2. Weight mg = 10⁻³ × 10 = 10⁻² N.
3. tan θ = F/mg = 0.1, so θ = tan⁻¹(0.1).

Teacher's Note:
1. In equilibrium tan θ equals the ratio of horizontal force to weight.
2. Convert 0.3 cm to 3 × 10⁻³ m.

Answer: (b) \( \frac{2k\lambda_1\lambda_2}{R} \)

Explanation:
1. Field of an infinite line charge at distance R is E = 2kλ₁/R.
2. Force on a length l of the other wire is λ₂l E.
3. Force per unit length = 2kλ₁λ₂/R.

Teacher's Note:
1. The field of a long wire falls as 1/R, not 1/R².
2. Force per unit length = linear density × field.

Answer: (d) both (1) and (2)

Explanation:
1. Inside (r < R) the enclosed charge is Q(r/R)³, so E = kQr/R³.
2. Outside (r ≥ R) the sphere behaves like a point charge at the centre: E = kQ/r².
3. Both statements are correct.

Teacher's Note:
1. Gauss's law with a spherical Gaussian surface gives both results.
2. Field is continuous at r = R where both expressions equal kQ/R².

Answer: (a) \( Q/\varepsilon_0 \)

Explanation:
1. By Gauss's law the flux through a closed surface equals the enclosed charge over ε₀.
2. The enclosed charge is Q.
3. Flux = Q/ε₀, independent of the radius 2R.

Teacher's Note:
1. Flux depends only on the enclosed charge, not on the size or shape of the surface.
2. Do not multiply by area.

Answer: (b) 300 units

Explanation:
1. The area vector of a surface in the x–y plane is along the z axis: A = 100 k̂.
2. Φ = E · A = 3 × 100.
3. Φ = 300 units.

Teacher's Note:
1. Only the component of E normal to the surface contributes to flux.
2. Here that is the z component, 3.

Answer: (b) zero

Explanation:
1. The potential at A is V = kq/AB − kq/AC.
2. The triangle is isosceles, so AB = AC = a.
3. Therefore V = 0.

Teacher's Note:
1. A is on the perpendicular bisector of the line joining +q and −q.
2. Every point of this bisector is at zero potential.