Continuous Charge Distribution

Correct answer: A. Transverse components cancel by symmetry

Explanation: For every small charge element on the ring, there is a diametrically opposite element at the same distance from the axial observation point. Their transverse electric-field components have equal magnitude and opposite directions, so they cancel pairwise. The axial components point in the same axis-related direction and add. Hence the resultant field is axial. Options B and C are false because the ring may carry nonzero charge, and D is false because the axial distance need not be zero.

Correct answer: A. Add them with their signs

Explanation: Total charge is obtained by integrating the signed surface charge density over the complete surface: Q = ∫σ dA. A positive-density region contributes positive charge, while a negative-density region contributes negative charge. These contributions may partially or completely cancel, so their signs must be retained. Taking only one sign would calculate only a partial contribution, not the total charge, and setting all densities to zero has no physical justification.

Correct answer: B. Positive and negative parts can balance each other

Explanation: The net charge is the signed sum or integral of all elemental charges, Q = ∫dq. Thus a distribution can contain positive and negative regions whose contributions cancel, giving Q = 0 even though individual elements carry nonzero charge. This is different from saying the charge density is zero everywhere. Therefore option B is possible; options A, C, and D incorrectly confuse zero net charge with the absence of local charge or with an impossible distribution.

Correct answer: A. The field is directed toward the negative charge

Explanation: A negative charge density means that the corresponding infinitesimal elements carry negative charge. By the definition of electric field, the force on a positive test charge due to a negative source charge is attractive; hence the elemental field points toward the negative element. The resultant field still depends on vector addition and geometry, so it is not necessarily simply inward everywhere or zero. The sign of density must therefore be included.

Correct answer: A. Because the distances and directions of small charge parts change

Explanation: The electric field is the vector sum or integral of contributions from all small charge elements. Changing the shape changes the positions of those elements relative to the observation point, so both their distances and the directions of their field vectors can change. Since each contribution depends on distance, approximately as 1/r² for a point element, and on direction, the resultant field can change substantially. Shape does not erase charge or automatically make density zero.

Correct answer: A. To add the contribution of every element and find the total effect

Explanation: A continuously charged wire cannot usually be treated as one point charge because different parts are at different positions relative to the observation point. We divide it into differential elements dq, calculate the field or potential produced by each element, and integrate or sum all contributions. Thus option A is correct; the other choices have no role in the mathematical treatment of a continuous charge distribution.

Correct answer: A. By multiplying the appropriate charge density by the corresponding small measure

Explanation: The differential charge depends on the type of distribution. For a line, dq = λ dl; for a surface, dq = σ dA; and for a volume, dq = ρ dV. Thus the appropriate density must be multiplied by the matching small length, area, or volume. Option A expresses this general rule. The element is not automatically zero, and mass or colour has no connection with charge calculation.

Correct answer: A. It remains the same

Explanation: For a uniform volume distribution, total charge is Q = ρV. Let the original values be ρ and V, so Q = ρV. After the changes, ρ′ = 2ρ and V′ = V/2. Therefore Q′ = ρ′V′ = (2ρ)(V/2) = ρV = Q. The increase in density exactly compensates for the decrease in volume, so option A is correct.

Correct answer: A. First identify whether the charge is spread along a line, over a surface, or through a volume

Explanation: The governing concept is that charge density depends on the geometrical region over which charge is distributed. For a line, use λ = dQ/dl; for a surface, use σ = dQ/dA; and for a volume, use ρ = dQ/dV. Identifying the distribution first selects both the correct density and integration element. Therefore option A is correct. Colour, memorization, and mass do not determine the electrostatic density formula.

Correct answer: A. It becomes half

Explanation: The governing relation for a uniform surface distribution is Q = σA, where Q is total charge, σ is surface charge density, and A is area. Since σ remains constant, changing the area directly changes Q in the same ratio. If the new area is A/2, then Q′ = σ(A/2) = Q/2. Thus option A is correct. Option B reverses the ratio, while C would be true only if the density changed inversely.

Correct answer: B. It doubles

Explanation: For a uniform volume charge distribution, the governing equation is Q = ρV, where ρ is volume charge density and V is volume. Because ρ is unchanged, total charge is directly proportional to volume. If the volume changes from V to 2V, then Q′ = ρ(2V) = 2Q. Therefore option B is correct. Option A would describe a decrease in volume, and option C ignores the direct proportionality.

Correct answer: C. It doubles

Explanation: Linear charge density is defined as λ = Q/L, where Q is total charge and L is the length. The charge is unchanged, but the new length is L/2. Hence λ′ = Q/(L/2) = 2Q/L = 2λ. Therefore the linear density doubles, so option C is correct. It does not remain constant because the same charge is concentrated into half the length; option A has the inverse trend.

Correct answer: A. It becomes half

Explanation: Surface charge density is defined by σ = Q/A. Here Q remains fixed while the area changes from A to 2A. Therefore σ′ = Q/(2A) = σ/2. The same total charge is spread over twice as much surface, so the charge per unit area is halved. Thus option A is correct. Option B incorrectly treats density as directly proportional to area, while option C would require the charge to double as well.

Correct answer: C. It becomes one fourth

Explanation: Volume charge density is defined as ρ = Q/V. Since the total charge Q is fixed and the volume changes from V to 4V, the new density is ρ′ = Q/(4V) = ρ/4. The same charge is distributed through four times the volume, so charge per unit volume becomes one fourth. Therefore option C is correct. Options A and B confuse the density with the volume, and D ignores the inverse relationship.

Correct answer: C. Two thirds

Explanation: For a uniform linear distribution, the linear charge density λ is constant and charge is given by Q = λL. Thus the charge in any selected segment is proportional to its length. If the selected length is (2/3)L, its charge is q = λ(2L/3) = 2Q/3. Therefore option C is correct. One third would correspond to one third of the length, while the whole charge requires the entire length.

Correct answer: A. One fifth

Explanation: In a uniform surface distribution, surface charge density σ is constant, and charge on an area is q = σA. Consequently, the charge fraction equals the area fraction. For a selected area A/5, q = σ(A/5) = Q/5. Hence option A is correct. One fourth and one half do not match the stated area ratio, and five times is impossible for a smaller portion of the same uniformly charged surface.

Correct answer: C. Three fourths

Explanation: For a uniform volume distribution, the volume charge density ρ is constant, so charge in a region is q = ρV. The charge fraction therefore equals the volume fraction. Selecting (3/4)V gives q = ρ(3V/4) = 3Q/4. Thus option C is correct. One fourth and one half are different volume fractions, while the whole charge would require selecting the complete volume.

Correct answer: B. Non-uniform linear density

Explanation: Linear charge density is charge per unit length, λ = Q/L. If two segments have equal lengths and equal λ, they must contain equal charges. Here the lengths are equal but the charges differ, so Q/L is different for the two segments. Therefore the linear charge density varies along the wire and the distribution is non-uniform; option B is correct. Surface and volume densities describe area and volume distributions, not this one-dimensional comparison.

Correct answer: A. Uniform surface density

Explanation: Surface charge density is defined as σ = Q/A, or charge per unit area. When two regions have equal areas and carry equal charges, their Q/A values are equal. This indicates that the surface density is constant between those regions, so the distribution is uniform. Therefore option A is correct. Unequal density would require unequal charge on equal areas, while volume density applies to charge distributed through a three-dimensional volume.

Correct answer: B. Non-uniform volume distribution

Explanation: The relevant concept is volume charge density, defined as charge per unit volume. If equal volumes contain unequal charges, their densities are unequal, so the charge density changes from place to place. Therefore the distribution is non-uniform in volume, making option B correct. Option A would require equal charge in equal volumes, while options C and D describe charge associated with length or surface area.

Correct answer: A. Zero

Explanation: Use the symmetry of a uniformly charged ring. For every small charge element on the ring, an element diametrically opposite to it is at the same distance from the centre and produces an electric field of equal magnitude in the opposite direction. These vector contributions cancel in pairs, so the resultant field at the centre is zero. Hence option A is correct; the field is not infinite or automatically negative.

Correct answer: B. Component along the axis

Explanation: Consider pairs of equal charge elements located symmetrically on opposite sides of the ring. At an axial point, the components perpendicular to the axis are equal in magnitude and opposite in direction, so they cancel. Their components parallel to the axis point in the same direction and add. Thus only the axial component remains, making option B correct; no tangential-only or complete cancellation occurs away from the centre.

Correct answer: A. Because opposite small parts are at equal distances

Explanation: The cancellation follows from geometric symmetry, not from the rod being neutral. For every small charge element on one side of the rod, there is an equal element at the same distance on the other side. Their horizontal field components have equal magnitudes but opposite directions, so the horizontal parts cancel. The components along the perpendicular bisector reinforce each other. Therefore option A is correct.

Correct answer: B. Away from the element

Explanation: A sufficiently small positive charge element is treated as a point-like source when finding its contribution to the total field. Electric field lines originate from positive charge, so the field produced by that element is directed radially outward, away from the element. Hence option B is correct. The field is not always zero, and circular direction is not a general property of an electrostatic point charge.

Correct answer: B. Toward the element

Explanation: The electric field direction is defined as the force direction on a positive test charge. A negative charge attracts such a test charge, so the field produced by a small negative element points toward that element. Therefore option B is correct. The field is not necessarily along an axis, and it is not always zero; its exact direction also depends on the element’s position relative to the observation point.