Continuous Charge Distribution

Correct answer: A. 1:1

Explanation: For a volume charge distribution, total charge is Q = ρV, where ρ is volume charge density and V is volume. If ρ₁ = 2ρ₂ and V₁ = V₂/2, then Q₁ = (2ρ₂)(V₂/2) = ρ₂V₂ = Q₂. Thus the two solids carry equal total charge. Option B would ignore the volume change, while option C reverses the comparison. Option A is correct.

Correct answer: B. Complete circular symmetry

Explanation: The key concept is superposition together with complete circular symmetry. For every small charge element on a uniformly charged ring, an opposite element produces an electric field of equal magnitude in the opposite direction at the centre. All such contributions cancel pairwise, even though the total charge is not zero. Therefore option B is correct; mass is irrelevant and non-uniformity would generally spoil the cancellation.

Correct answer: B. The field may be non-zero because symmetry is broken.

Explanation: For a uniformly charged ring, opposite elements make equal and opposite contributions at the centre, so the field cancels. Non-uniform charge density removes that pairwise equality; one part may contribute more strongly than its opposite. The resultant field can therefore be non-zero, although it could accidentally vanish for a specially balanced distribution. Thus option B is the only generally correct statement.

Correct answer: A. The rod’s finite size becomes negligible compared with the observation distance.

Explanation: The far-field or point-charge approximation applies when the observation distance r is much larger than the rod length L. Then the relative variation of distance from different elements, roughly L/r, is small, so their individual field contributions have nearly the same distance dependence. The field is therefore well approximated by that of the total charge concentrated at a representative point. The charge and density do not disappear, so option A is correct.

Correct answer: B. Toward the element

Explanation: A small surface element carries charge dq = σ dA. Since dA is positive and σ is negative, dq is negative. The electric field of a negative point-like element is directed toward that element, along the line joining the observation point to the charge. The exact overall field can depend on all elements, but the contribution of this negative element points inward. Hence option B is correct.

Correct answer: A. Because the unit shows whether charge is spread over length, area, or volume

Explanation: The unit identifies the geometrical model of a continuous charge distribution. Charge per length, λ = Q/L, has unit C m⁻¹ and represents a linear distribution. Charge per area, σ = Q/A, has unit C m⁻² and represents a surface distribution, while charge per volume, ρ = Q/V, has unit C m⁻³ and represents a volume distribution. Therefore the unit helps select the correct formula and integration element.

Correct answer: B. The average charge per unit length over the whole wire

Explanation: For a wire of total length L and total charge Q, the average linear charge density is λ_avg = Q/L. It gives the charge per unit length averaged over the entire wire, not the value at each location. In a non-uniform distribution, the local density λ(x) changes with position, so a small segment has charge dq = λ(x)dx. Thus only option B follows from the average.

Correct answer: B. Both will be equal in magnitude but opposite in sign

Explanation: For a surface, total charge is Q = σA. Let the two equal areas be A, with surface charge densities +σ and −σ of equal magnitude. Then Q₁ = (+σ)A and Q₂ = (−σ)A, so |Q₁| = |Q₂| but Q₁ = −Q₂. Therefore the charges are equal in magnitude and opposite in sign. Equal signs, zero charge, or infinite charge do not follow from the stated conditions.

Correct answer: A. A point charge with the same total charge

Explanation: At an axial distance much larger than the ring radius, the ring’s dimensions are small compared with the observation distance. Its detailed shape then has little effect on the leading electric field, and the total charge Q dominates. The far-field potential approaches V ≈ (1/4πε₀)Q/r, the same leading form as a point charge carrying Q. Thus option A is correct; it does not become an infinite line or an uncharged object.

Correct answer: A. Volume charge density changes with position

Explanation: Volume charge density is defined as ρ = dQ/dV, the charge contained per unit volume. In a non-uniform distribution, ρ is a function of position, such as ρ(x,y,z), rather than a single constant. Consequently, two equal volume elements can contain different charges because dQ = ρ dV and their local values of ρ differ. The other choices contradict the meaning of volume or charge density and do not explain the observation.

Correct answer: B. Four times

Explanation: For a uniform surface charge distribution, total charge is given by Q = σA, where σ is surface charge density and A is area. If σ changes to 2σ and A changes to 2A, the new charge is Q′ = (2σ)(2A) = 4σA = 4Q. Therefore option B is correct. Option A accounts for only one doubled factor and ignores the second.

Correct answer: B. Double

Explanation: For a uniform line-charge distribution, total charge is Q = λL, where λ is linear charge density and L is length. After the changes, λ′ = λ/2 and L′ = 4L. Hence Q′ = (λ/2)(4L) = 2λL = 2Q. Therefore the total charge doubles and option B is correct. The half and four-times choices consider only one factor at a time.

Correct answer: A. Three times

Explanation: For a uniform volume charge distribution, total charge is Q = ρV, with ρ denoting volume charge density and V denoting volume. The changed values are ρ′ = ρ/3 and V′ = 9V. Thus Q′ = (ρ/3)(9V) = 3ρV = 3Q. Therefore option A is correct. The other numerical choices omit or misapply one of the two factors.

Correct answer: A. Length is a geometrical measure and is positive

Explanation: Length is a geometric magnitude: for a physical segment its value is non-negative and is positive when the segment has non-zero extent. Total charge, however, may be positive or negative depending on the sign of the charge or linear density, since Q = λL. Therefore option A gives the correct reason. Option B is false because negative charge is physically possible.

Correct answer: A. Positive and negative parts may balance each other

Explanation: The governing idea is signed integration: total charge is Q = ∫ρ dV, or the corresponding line or surface integral. A zero value of Q means that positive and negative contributions cancel in the total, not that every local value of density is zero. Therefore, option A is correct. Option C confuses net charge with local density, while option D is also unjustified because a charge distribution with zero net charge can still produce a non-zero electric field.

Correct answer: A. Decide whether charge is distributed along a line, surface, or volume, and whether the density is uniform or non-uniform

Explanation: The governing concept is choosing the correct charge element before calculating. A line distribution uses dq = λ dl, a surface distribution uses dq = σ dA, and a volume distribution uses dq = ρ dV. Uniformity decides whether the density can be taken outside the integral. Hence option A is correct. The colour, temperature, mass, sound, or memorised total charge does not determine the appropriate electrostatic integration method.

Correct answer: A. To understand total charge and electric field by adding contributions from small elements of a spread-out charge

Explanation: The governing principle is superposition applied to an extended distribution. A small element carries dq, and its contribution is summed continuously: Q = ∫dq; similarly, the electric field is obtained from dE and integrated over the line, surface, or volume. Thus option A expresses the purpose correctly. Option B reverses the idea, because the charge is spread out; option C is false because electric field is a vector; option D has no physical basis.

Correct answer: C. 12 coulomb per metre

Explanation: For a uniform linear distribution, the governing relation is λ = Q/L. After the change, Q remains 48 C and the new length is L = 4 m. Therefore, λnew = 48/4 = 12 C m⁻¹. Option C is correct. The old density would be 48/12 = 4 C m⁻¹, so option A is the value before shortening, not the required new value; the other options do not satisfy Q = λL.

Correct answer: B. 12 coulomb per square metre

Explanation: The governing relation for a uniform surface distribution is σ = Q/A. With unchanged charge Q = 60 C and new area A = 5 m², the new density is σnew = 60/5 = 12 C m⁻². Therefore, option B is correct. Using the original area gives 60/20 = 3 C m⁻², which explains option A as the initial density, not the final one. The remaining values are inconsistent with Q = σA.

Correct answer: B. It becomes one and a half times

Explanation: The governing relation is Q = λL for a uniform linear distribution. If λ changes to 3λ and L changes to L/2, then Qnew = (3λ)(L/2) = 3Q/2. Thus the new charge is one and a half times the original charge, making option B correct. Option C ignores the simultaneous length reduction, while option D multiplies three and two instead of using the factor one-half.

Correct answer: B. It becomes double

Explanation: For a uniform surface distribution, total charge is Q = σA. The changed charge is Qnew = (σ/4)(8A) = 2σA = 2Q. Therefore, option B is correct: the total charge doubles. Option A would result if the area remained unchanged, and option D would result if the density remained unchanged. Both changes must be included in the product, so neither of those distractors is valid.

Correct answer: A. It remains the same

Explanation: The governing relation for a uniform volume distribution is Q = ρV. After the changes, Qnew = (5ρ)(V/5) = ρV = Q. Therefore, option A is correct and the total charge remains unchanged. Option B considers only the density factor, option C considers only the volume factor, and option D multiplies the numerical changes without recognising that one factor is the reciprocal of the other.

Correct answer: B. Twenty to five

Explanation: The governing relation for a surface charge distribution is Q = σA, where σ is surface charge density and A is area. Therefore, Q₁:Q₂ = (σ₁A₁):(σ₂A₂) = (2×10):(5×1) = 20:5 = 4:1. Thus option B is correct. Option A is not the resulting product ratio, while C ignores area and D ignores density.

Correct answer: C. Two to one

Explanation: For a volume charge distribution, total charge is Q = ρV, where ρ is volume charge density and V is volume. Hence Q₁:Q₂ = (3×8):(4×3) = 24:12 = 2:1. Therefore option C is correct. Option B results from an incorrect multiplication, while option D copies only the density ratio and ignores the different volumes.

Correct answer: B. The field may become non-zero because symmetry is broken

Explanation: The complete uniformly charged ring has rotational symmetry: every charge element has an opposite partner whose electric-field contribution cancels it at the centre. Removing even a small part removes that matching contribution, so the cancellation is incomplete and a non-zero resultant field can appear. Therefore option B is correct; the other choices contradict charge conservation or confuse surface and volume densities.