JEE Physics Electric Charges and Fields MCQs Set 22

Answer: (b) \( \epsilon_0E_0a^3 \)

Explanation:
1. Only the two faces perpendicular to x contribute: flux out at x = a is \( E_0a\cdot a^2 \) and at x = 0 it is 0.
2. Net flux \( \phi=E_0a^3 \), so \( q=\epsilon_0\phi=\epsilon_0E_0a^3 \).

Teacher's Note:
1. The field is parallel to the x-axis, so the side faces carry no flux.
2. Gauss's law: charge = epsilon-zero times net outward flux.

Answer: (a) 173.2

Explanation:
1. The area vector is \( 100\hat k \).
2. \( \phi=\vec E\cdot\vec A=100\sqrt3=173.2\ \text{V m} \).

Teacher's Note:
1. Only the component of E normal to the plane matters.
2. Here that is the k component.

Answer: (d) aA

Explanation:
1. The area vector of a surface in the yz plane is \( A\hat i \).
2. \( \phi=\vec E\cdot\vec A=aA \).

Teacher's Note:
1. Normal of the yz plane is along x.
2. The j component is parallel to the surface and gives no flux.

Answer: (b) \( \dfrac{2\lambda\sqrt{R^2-y^2}}{\epsilon_0} \)

Explanation:
1. The length of the line inside the sphere is the chord \( 2\sqrt{R^2-y^2} \).
2. Charge enclosed \( q=2\lambda\sqrt{R^2-y^2} \), and \( \phi=q/\epsilon_0 \).

Teacher's Note:
1. Only the enclosed part of the line matters.
2. Use Pythagoras for the half-chord.

Answer: (a) \( \dfrac{q}{20\epsilon_0} \)

Explanation:
1. Area of the sphere \( 4\pi R^2=4\pi\cdot\dfrac1\pi=4\ \text{m}^2 \).
2. The flux is spread uniformly: \( \phi=\dfrac{q}{\epsilon_0}\cdot\dfrac{0.2}{4}=\dfrac{q}{20\epsilon_0} \).

Teacher's Note:
1. Uniform distribution means flux is proportional to area.
2. Check that 4πR² = 4 first.

Answer: (b) \( \dfrac{\sqrt{l^2+b^2+h^2}}{h} \)

Explanation:
1. Flux = charge inside over ε₀, and charge inside is λ times the length of wire within the cuboid.
2. The longest possible length is the space diagonal \( \sqrt{l^2+b^2+h^2} \); the shortest is the smallest edge h.
3. Ratio \( =\dfrac{\sqrt{l^2+b^2+h^2}}{h} \).

Teacher's Note:
1. A straight wire passing through the body has maximum length along the longest diagonal.
2. Minimum length corresponds to passing perpendicularly through the thinnest dimension.

Answer: (b) \( \dfrac{\pi(R^2-x^2)\sigma}{\epsilon_0} \)

Explanation:
1. The sheet inside the sphere is a disc of radius \( \sqrt{R^2-x^2} \).
2. Enclosed charge \( q=\sigma\pi(R^2-x^2) \), so \( \phi=\dfrac{\sigma\pi(R^2-x^2)}{\epsilon_0} \).

Teacher's Note:
1. The intersection of a plane and a sphere is a circle.
2. Gauss's law uses only the enclosed charge.

Answer: (b) \( \dfrac{\sqrt3\,\lambda a}{\epsilon_0} \)

Explanation:
1. Flux = (charge inside)/ε₀ = λ × (length of string inside)/ε₀.
2. The longest chord of the cube is its space diagonal \( a\sqrt3 \).
3. So \( \phi_{max}=\sqrt3\,\lambda a/\epsilon_0 \).

Teacher's Note:
1. The longest straight line inside a cube is the body diagonal.
2. The shortest chord is an edge perpendicular passage, length a.

Answer: (d) 80

Explanation:
1. \( \phi=EA\cos\theta=20\times8\times\cos60^\circ=80 \).

Teacher's Note:
1. The angle is measured from the normal to the surface, not the surface itself.
2. Area = 2 × 4 = 8 square metres.

Answer: (c) 0

Explanation:
1. Net charge \( =(2-5-3+6)\times10^{-6}=0 \).
2. Flux = net enclosed charge divided by ε₀ = 0.

Teacher's Note:
1. Add the charges with signs.
2. Gauss's law involves the sum of enclosed charge.

Answer: (a) (i) and (iii) only

Explanation:
1. The field points inward, so the flux through a sphere about the origin is negative: the enclosed charge is negative. (i) is true; (ii) fails because a sphere far from the origin need not enclose negative charge.
2. Magnitude: \( |q|=\epsilon_0\cdot E\cdot4\pi r^2=8.85\times10^{-12}\times100\,r\cdot4\pi r^2 \) with r = 0.03 m gives \( 3\times10^{-13}\ \text{C} \), so (iii) is true and (iv) false.

Teacher's Note:
1. Gauss's law: inward field means negative enclosed charge.
2. Substitute r = 0.03 m carefully.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. The flux depends only on Q₁, but the field E at points of S depends on all charges.
2. (i) true; (ii) true; (iii) true since an outside charge gives a field but zero net flux.
3. (iv) is false: with Q₁ ≠ 0 the field on S cannot vanish everywhere.

Teacher's Note:
1. Distinguish the field (all charges) from the flux (enclosed charge only).
2. Outside charges contribute zero net flux but a non-zero field.

Answer: (a) (iii) and (iv) only

Explanation:
1. Gauss's law is true for any charge distribution and any medium (with the appropriate permittivity), so (i) and (ii) are false.
2. The field E in the integral is the total field, and charges outside give zero net flux, so (iii) and (iv) are true.

Teacher's Note:
1. Gauss's law is a general law; symmetry is only needed to apply it easily.
2. Outside charges cancel in the net flux.

Answer: (a) (iii) and (iv) only

Explanation:
1. The conductor's cavity surface is an equipotential, so (iii) is true.
2. Gauss's law over the cavity surface encloses q, so (iv) is true.
3. The surface is ellipsoidal and its points are at different distances from q, so the field and charge density differ: (i) and (ii) are false.

Teacher's Note:
1. The induced charge density follows the local field, which depends on distance from q.
2. Conductor surfaces are equipotentials.

Answer: (a) (i) and (iv) only

Explanation:
1. The net enclosed charge is zero, so the flux is zero: (i) true.
2. A dipole's field is nowhere zero on the sphere, so (ii) is false.
3. The potential \( kp\cos\theta/r^2 \) vanishes only on the equatorial circle, so (iii) false and (iv) true.

Teacher's Note:
1. Potential of a dipole depends on the angle from the axis.
2. Net charge zero means net flux zero.

Answer: (a) volt metre

Explanation:
1. \( \phi=\vec E\cdot\vec A \): \( \text{N/C}\times\text{m}^2=\text{N m}^2/\text{C} \) = (V/m) × m² = V m.

Teacher's Note:
1. Flux = field times area.
2. Electric field can be expressed in V/m.

Answer: (a) 2

Explanation:
1. Flux through the face x = 1: \( -4 \); face x = 0: 0. Face y = 1: \( +6 \); face y = 0: 0. Faces perpendicular to z: 0.
2. Net flux \( -4+6=2 \) so \( q=2\epsilon_0 \) and α = 2.

Teacher's Note:
1. Take the outward normal for every face.
2. Field components along z are zero.

Answer: (b) \( \dfrac{q}{2\epsilon_0} \)

Explanation:
1. When the charge is very close compared with the size of the square, the square looks like an infinite plane.
2. Half of the flux lines pass through the plane: \( \phi=\dfrac{q}{2\epsilon_0} \).

Teacher's Note:
1. Think of the square as an infinite plane with the charge just in front of it.
2. An infinite plane intercepts exactly half of the total flux.

Answer: (d) \( \dfrac{Q}{2\epsilon_0} \)

Explanation:
1. Flux equals the charge inside divided by ε₀; the rod starts at the cube's centre.
2. The smallest length of rod inside the cube is when the rod runs along a line perpendicular to a face, leaving the cube after l/2.
3. Charge inside \( Q/2 \), flux \( \dfrac{Q}{2\epsilon_0} \).

Teacher's Note:
1. Other orientations leave through an edge or a corner, a longer distance from the centre.
2. Charge is proportional to length for a uniform rod.

Answer: (a) \( \dfrac{aV}{b} \)

Explanation:
1. Initial charge \( Q=4\pi\epsilon_0aV \).
2. After connection all the charge moves to B, whose potential is \( \dfrac{kQ}{b}=\dfrac{aV}{b} \) (A sits at the same potential).

Teacher's Note:
1. Charge in contact with the inner wall of a hollow conductor goes to the outer surface.
2. The new potential is total charge over b.

Answer: (a) \( 2RlE \)

Explanation:
1. The flux entering the plane face ABCD (area \( 2R\cdot l \), normal to the field) must leave through the curved surface, as there is no charge inside.
2. So the flux through the curved surface has magnitude \( 2RlE \).

Teacher's Note:
1. The curved surface subtends the same projected area as the plane face.
2. Flux through a closed surface with no charge is zero.

Answer: (a) \( 4\pi aR \)

Explanation:
1. This is the field of a long line charge along the z-axis: \( E=\dfrac{\lambda}{2\pi\epsilon_0\rho}=\dfrac{a}{\rho} \), so \( \lambda=2\pi\epsilon_0a \).
2. The length of the line inside the sphere is 2R, so \( q=2R\lambda=4\pi\epsilon_0aR \).
3. Flux \( =q/\epsilon_0=4\pi aR \).

Teacher's Note:
1. The field has no z-component and falls as 1/ρ, the signature of a line charge.
2. Use Gauss's law with the enclosed length.

Answer: (b) \( 10^5\ \text{V m} \)

Explanation:
1. The ring subtends a cone of half-angle θ with \( \cos\theta=\dfrac{0.4}{\sqrt{0.4^2+0.3^2}}=0.8 \).
2. Flux through a cap: \( \phi=\dfrac{q}{2\epsilon_0}(1-\cos\theta)=\dfrac{q}{2\epsilon_0}\times0.2=\dfrac{q}{10\epsilon_0} \).
3. So \( \phi=\dfrac{8.85\times10^{-6}}{10\times8.85\times10^{-12}}=10^5\ \text{V m} \).

Teacher's Note:
1. The flux through a disc equals charge over 2ε₀ times (1 − cos of the half-angle).
2. The flux equals the solid-angle fraction times q/ε₀.