JEE Physics Electric Charges and Fields MCQs Set 06

Answer: (c) \( V_A = V_C,\ V_B \lt V_D \)

Explanation:
1. Potential falls in the direction of the field. The field points to the right, so the right-hand point B is at lower potential than the left-hand point D: V_B < V_D.
2. A and C lie on the same vertical line, perpendicular to the field, so they are on the same equipotential: V_A = V_C.

Teacher's Note:
1. Equipotential surfaces are perpendicular to field lines.
2. Moving along the field, potential decreases.

Answer: (c) 5 V/m

Explanation:
1. From x = 12 m to x = 16 m the graph is a straight line from 30 V to 10 V.
2. Slope dV/dx = (10 − 30)/(16 − 12) = −5 V/m.
3. E = −dV/dx = +5 V/m.

Teacher's Note:
1. Field is the negative gradient of potential.
2. x = 13 m lies on the last straight segment, 12 m to 16 m.

Answer: (a) \( \text{C}^2\,\text{N}^{-1}\text{m}^{-2} \)

Explanation:
1. Coulomb's law: F = q₁q₂/(4πε₀r²).
2. So ε₀ = q₁q₂/(4πFr²), with unit C²/(N m²).
3. The unit is C² N⁻¹ m⁻².

Teacher's Note:
1. k = 1/(4πε₀) has unit N m² C⁻²; ε₀ has the reciprocal unit.
2. Always rearrange Coulomb's law to find units.

Answer: (c) They are radial for a point charge

Explanation:
1. Lines start on positive charge and end on negative charge, so (1) is wrong.
2. Dense lines mean a strong field, so (2) is wrong.
3. Field lines are only a pictorial aid, so (4) is wrong.
4. For an isolated point charge the lines are straight and radial: (3).

Teacher's Note:
1. Line density measures field strength.
2. Field lines are imaginary.

Answer: (a) \( 4\phi \)

Explanation:
1. Flux is proportional to the enclosed charge: φ = 20 µC/ε₀.
2. 80 µC gives 4φ.
3. The change in flux due to adding 80 µC is 4φ (the total becomes 5φ).

Teacher's Note:
1. Gauss's law is linear in charge.
2. Read carefully whether the question asks for the change or the total.

Answer: (b) \( 100Q/\varepsilon_0 \)

Explanation:
1. Length of wire inside the cylinder = 1 m = 100 cm.
2. Enclosed charge = 100 Q.
3. By Gauss's law, Φ = 100Q/ε₀.

Teacher's Note:
1. Check the unit of charge density: per cm, not per metre.
2. Flux through the whole closed surface equals enclosed charge over ε₀.

Answer: (b) the number of incoming and outgoing electric lines are equal

Explanation:
1. Zero enclosed charge means zero net flux.
2. Net flux zero means as many lines leave the surface as enter it.
3. E need not be zero at each point; outside charges can produce a field.

Teacher's Note:
1. Zero net flux does not mean zero field.
2. Flux counts lines in minus lines out.

Answer: (c) 120 V/m

Explanation:
1. Outside: E = kQ/r², so kQ = 100 × (0.2)² = 4 SI units.
2. Inside, E = kQr/R³ = 4 × 0.03/(0.1)³.
3. E = 0.12/0.001 = 120 V/m.

Teacher's Note:
1. Inside a uniformly charged solid sphere E grows linearly with r.
2. Find kQ from the outside data first.

Answer: (a) 10

Explanation:
1. A surface in the YZ plane has its normal along x, so A = 2 î.
2. Φ = E·A = 5 × 2 = 10 N m²/C.

Teacher's Note:
1. Only the component of E along the normal counts.
2. Normal to the YZ plane is the x axis.

Answer: (d) \( q/6\varepsilon_0 \)

Explanation:
1. Imagine a cube of side a with the charge at its centre: the square is one face.
2. Total flux q/ε₀ divides equally over 6 faces.
3. Flux through the square = q/6ε₀.

Teacher's Note:
1. A charge a/2 above the centre of a square of side a sits at the centre of a cube.
2. Use symmetry: six equal faces.

Answer: (b) the electric potential is zero

Explanation:
1. Each point on the bisector is equidistant from the two charges, so the potentials cancel: V = 0.
2. The field is not zero; it is parallel to the line joining the charges.

Teacher's Note:
1. The bisector is an equipotential surface.
2. Fields of equal and opposite charges add along the joining line.

Answer: (a) \( \frac{1}{4\pi\varepsilon_0}\frac{Q}{R} \)

Explanation:
1. Inside a conductor E = 0, so the potential is constant and equal to the surface value.
2. V = kQ/R.

Teacher's Note:
1. Zero field does not mean zero potential.
2. Potential inside equals potential on the surface.

Answer: (a) \( V \)

Explanation:
1. Potential difference between the sphere and the shell is set by the field in the region between them and beyond.
2. Adding charge to the shell shifts the potential of the shell and everything inside it by the same amount.
3. The difference between the two surfaces is unchanged: V.

Teacher's Note:
1. A charge on the outer shell changes the potential of the inner sphere by the same amount as that of the shell.
2. Only the enclosed charge Q affects the field between the surfaces.

Answer: (b) E ≠ 0, V = 0

Explanation:
1. All corners are at the same distance d from the centre.
2. V = k(2 − 3 − 4 + 5)/d = 0.
3. The fields do not cancel: opposite corners carry 2C, −4C and −3C, 5C, whose fields are unequal, so E ≠ 0.

Teacher's Note:
1. Potential is scalar and adds algebraically; the field is a vector.
2. Check the opposite pairs: unequal charges on a diagonal leave a net field.

Answer: (a) 2 V

Explanation:
1. The integral from infinity to the centre of −E·dl is the potential at the centre minus the potential at infinity.
2. Potential at the centre depends only on total charge and radius: V = kQ/R = 9 × 10⁹ × 1.11 × 10⁻¹⁰/0.5 = 2 V.
3. The non-uniformity does not matter because every element is at distance R.

Teacher's Note:
1. V = −∫E·dl from infinity to the point.
2. Every charge element is at the same distance R from the centre.

Answer: (d) zero

Explanation:
1. Inside a charged hollow conductor the electric field is zero everywhere.
2. The potential is constant at 10 V but the field is zero.

Teacher's Note:
1. Field is zero inside a conductor; potential is not.
2. Do not divide V by R here.

Answer: (c) \( \frac{Q}{4\pi\varepsilon_0 R} \)

Explanation:
1. A point inside a conductor is at the same potential as its surface.
2. V = Q/(4πε₀R).

Teacher's Note:
1. The point R/2 is inside the sphere, so use the surface potential.
2. Do not use kQ/r with r = R/2.

Answer: (a) 5 V

Explanation:
1. V(10) − V(20) = ∫ from 10 to 20 of E dx = 100(1/10 − 1/20).
2. = 100 × 0.05 = 5 V.

Teacher's Note:
1. Integrate E from the first point to the second to find the potential drop.
2. ∫dx/x² = −1/x.

Answer: (b) \( V_A > V_B \)

Explanation:
1. Potential decreases along the field direction, so V_A > V_B (B is further along +x).
2. C lies on the same equipotential plane as A (x = 0), so V_A = V_C.

Teacher's Note:
1. AC is perpendicular to E, so no work is done along AC.
2. Potential decreases in the direction of the field.

Answer: (d) zero

Explanation:
1. E = −dV/dx = −(10x − 10).
2. At x = 1 m, E = −(10 − 10) = 0.

Teacher's Note:
1. Field is the negative gradient of potential.
2. Differentiate first, then substitute.

Answer: (c) 0.2 µC

Explanation:
1. V/E = (kq/r)/(kq/r²) = r = 600/200 = 3 m.
2. q = Vr/k = 600 × 3/(9 × 10⁹) = 2 × 10⁻⁷ C = 0.2 µC.

Teacher's Note:
1. Dividing V by E gives the distance.
2. Then use V = kq/r.

Answer: (c) \( \frac{\sigma}{\varepsilon_0}(R+r) \)

Explanation:
1. Charges are Q₁ = 4πR²σ and Q₂ = 4πr²σ.
2. Potential at the centre V = kQ₁/R + kQ₂/r = σR/ε₀ + σr/ε₀.
3. V = (σ/ε₀)(R + r).

Teacher's Note:
1. Potentials of the two shells simply add at the centre.
2. kQ/R with Q = 4πR²σ simplifies to σR/ε₀.

Answer: (b) the field is non-zero but the potential is zero

Explanation:
1. The centre is equidistant from all three charges, so V = k(2q − q − q)/r = 0.
2. The fields of −q and −q point towards those charges and add up to a field of magnitude kq/r² along the line away from the 2q vertex, in the same direction as the field of 2q (2kq/r²). Net field 3kq/r² ≠ 0.

Teacher's Note:
1. Potential is a scalar sum; the field is a vector sum.
2. Draw the three field vectors to check the directions.

Answer: (d) 20 N

Explanation:
1. Ex = −∂V/∂x = −(6 − 8y²) = −6 at the origin.
2. Ey = −∂V/∂y = −(−16xy − 8 + 6z) = 8 at the origin.
3. Ez = −∂V/∂z = −(6y − 8z) = 0.
4. |E| = √(36 + 64) = 10 N/C and F = qE = 2 × 10 = 20 N.

Teacher's Note:
1. E is the negative gradient of V in three dimensions.
2. Remember to multiply by the charge at the end.

Answer: (c) start from positive charge and end at negative charge

Explanation:
1. Field lines begin on positive charges and end on negative charges (or at infinity).
2. Two lines cannot cross, since the field has one direction at a point.
3. Lines are not closed curves for electrostatic fields.

Teacher's Note:
1. Crossing lines would give two field directions at one point.
2. Electrostatic field lines are not closed loops.