Continuous Charge Distribution

Correct answer: A. The longest part

Explanation: The governing relation for a continuous line charge is Q = λL, where λ is the uniform linear charge density and L is the length of the selected part. Since λ is identical for all three parts, their charges are proportional to their lengths. The longest part therefore contains the greatest total charge. The shortest part has less charge, while equal charges would occur only if the lengths were equal.

Correct answer: A. The larger-area part

Explanation: For a uniformly surface-charged plate, the total charge on a portion is Q = σA, where σ is the constant surface charge density and A is that portion’s area. Because σ is the same everywhere, charge increases directly with area. Hence the part with the larger area has greater total charge. Equal charges would require equal areas, so options B, C, and D do not fit the stated conditions.

Correct answer: A. The larger-volume part

Explanation: For a solid with uniform volume charge density, the charge in a small or finite portion is Q = ρV, where ρ is constant and V is the portion’s volume. Thus, when comparing portions of the same solid, the charge is directly proportional to volume. The larger-volume part contains more charge. A smaller part has less charge, and equal charges require equal volumes under uniform density.

Correct answer: B. A small area

Explanation: A density given in coulombs per square metre is surface charge density, written as σ. The charge on a small surface element is obtained from dq = σ dA, where dA is the small area element. Multiplying by a length or volume would give incompatible units, and time is unrelated to static charge distribution. Therefore the required factor is the small area.

Correct answer: C. A small volume

Explanation: Coulombs per cubic metre is the unit of volume charge density, denoted by ρ. For a small three-dimensional element, the differential charge is dq = ρ dV, where dV is its small volume. A length, area, or unspecified distance does not supply the required cubic-metre factor and gives incorrect dimensions. Hence the charge element must be calculated using the small volume.

Correct answer: A. A small length

Explanation: Charge density per metre is linear charge density, represented by λ and measured in C/m. For a small element of the charged line, the charge is dq = λ dl, where dl is the small length. This multiplication cancels metres and produces coulombs. Area and volume belong to surface and volume densities, while force is not a geometrical measure of the distribution, so option A is correct.

Correct answer: A. It remains the same

Explanation: The governing relation for a uniform surface charge distribution is Q = σA, where Q is total charge, σ is surface charge density, and A is area. If σ changes to 2σ and A changes to A/2, the new charge is Q′ = (2σ)(A/2) = σA = Q. Thus, the increase in density exactly compensates for the decrease in area. Option B would apply if only density doubled, while option C would apply if only area were halved.

Correct answer: A. It becomes three times

Explanation: For a uniform volume charge distribution, total charge is Q = ρV, where ρ is volume charge density and V is volume. After the changes, ρ′ = ρ/4 and V′ = 12V. Hence Q′ = (ρ/4)(12V) = 3ρV = 3Q. Therefore the total charge becomes three times its original value. It does not remain unchanged because the twelvefold increase in volume is larger than the fourfold reduction in density.

Correct answer: A. Three sevenths

Explanation: For a uniform linear charge distribution, the linear density λ is constant, so charge is directly proportional to length: Qpart = λLpart and Qtotal = λLtotal. Dividing these relations gives Qpart/Qtotal = Lpart/Ltotal. Since the stated length ratio is 3/7, the charge ratio is also 3/7. The remaining length carries the remaining 4/7 of the total charge.

Correct answer: B. Five twelfths

Explanation: In a uniform surface distribution, surface density σ is constant, so the charge on any part is Qpart = σApart, while total charge is Qtotal = σAtotal. Therefore Qpart/Qtotal = Apart/Atotal. The given area ratio is 5/12, so the charge fraction is also 5/12. The complementary area has 7/12 of the charge, which explains why option A is not correct for the specified part.

Correct answer: C. Nine tenths

Explanation: The governing concept is uniform volume charge density, defined as ρ = Q/V. Since ρ is constant, charge is directly proportional to volume: q/Q = v/V. Substituting v/V = 9/10 gives q/Q = 9/10. Therefore option C is correct. Option A is the complementary fraction, option B is unrelated, and option D would require the part to occupy the entire volume.

Correct answer: D. All are the same

Explanation: Linear charge density is defined by λ = Q/L, meaning charge per unit length. For a uniformly charged wire, λ has the same value at every location before cutting. Cutting only changes the total charge and length of each separated piece in the same proportion; it does not alter the local density. Thus every piece retains the original λ, so option D is correct. Length alone does not determine density.

Correct answer: C. All are the same

Explanation: Surface charge density is σ = Q/A, the charge per unit area. The statement that the original plate is uniformly charged means σ is constant across its surface. After division, a larger piece contains more total charge and a smaller piece contains less, but their charge-to-area ratios remain equal to the original σ. Therefore option C is correct; area changes total charge, not uniform density.

Correct answer: C. All are the same

Explanation: Volume charge density is ρ = Q/V. For a uniformly charged solid, every small region has the same charge per unit volume. When the solid is divided, the total charge in each piece changes in proportion to its volume: a larger piece has more charge, while a smaller one has less. Their ratios Q/V remain equal, so option C is correct. Division does not create infinite or unequal density.

Correct answer: B. Negative

Explanation: For a uniform surface charge distribution, the total charge is Q = σA. Area is a geometrical measure and is positive for a nonzero surface, whereas σ carries the charge sign. Consequently, if σ < 0 and A > 0, their product is Q < 0. Thus option B is correct. A positive answer reverses the sign, zero would require zero density or zero area, and “signless” is not a valid property of electric charge.

Correct answer: B. Negative

Explanation: Linear charge density is defined as charge per unit length: λ = Q/L for a uniform distribution. Length L is a positive geometrical quantity. Therefore a negative total charge Q divided by positive L gives λ < 0, so option B is correct. The density cannot be positive because that would require a positive numerator; it is not zero because Q is stated to be negative, and charge density is not signless.

Correct answer: A. The density will be zero

Explanation: For a uniform surface distribution, surface charge density is constant and is defined by σ = Q/A. Here Q = 0 while the area A is non-zero, so division is valid and gives σ = 0/A = 0. Therefore option A is correct. A positive or negative density would produce a non-zero total charge for a uniform surface of non-zero area, and an infinite density is neither implied nor mathematically obtained from these conditions.

Correct answer: A. Zero

Explanation: For a uniform linear charge distribution, total charge is given by Q = λL, where λ is the constant linear charge density and L is the length. Substituting λ = 0 and any finite non-zero length gives Q = 0 × L = 0. Thus option A is correct. The non-zero length does not create charge by itself; positive, negative, or infinite total charge would require a corresponding non-zero or undefined density, which is not stated.

Correct answer: B. −24 C

Explanation: The governing relation for a uniform volume charge distribution is Q = ρV, where ρ is volume charge density and V is volume. Substituting the values gives Q = (−6 C/m³)(4 m³) = −24 C. The cubic-metre units cancel, and the negative sign remains because the density represents negative charge. Therefore, option B is correct; option A reverses the sign, while C ignores the volume factor and D is unsupported.

Correct answer: B. −14 C

Explanation: For a uniform surface charge distribution, total charge is obtained from Q = σA, where σ is surface charge density and A is area. Here Q = (−7 C/m²)(2 m²) = −14 C. The area units cancel correctly, and the negative sign indicates that the distributed charge is negative. Thus option B is correct. Option A has the wrong sign, option C omits multiplication by area, and option D incorrectly assumes cancellation.

Correct answer: B. −5 C/m

Explanation: The governing relation for a uniformly charged wire is λ = Q/L, where λ is linear charge density, Q is total charge, and L is length. Substitution gives λ = (−60 C)/(12 m) = −5 C/m. Since length is positive, the density retains the charge sign. Therefore option B is correct; A changes the sign, C fails to divide by length, and D ignores the nonzero charge.

Correct answer: C. One-ninth

Explanation: The governing relation for a uniform surface distribution is σ = Q/A, where σ is surface charge density, Q is total charge, and A is area. Since Q remains constant while the new area is 9A, the new density is σ′ = Q/(9A) = σ/9. Thus the charge is spread more thinly, so option C is correct. It does not become three or nine times; those choices reverse the area-density relation.

Correct answer: C. One twenty-seventh

Explanation: For a uniform volume distribution, the governing definition is ρ = Q/V, where ρ is volume charge density, Q is total charge, and V is volume. With Q fixed and the new volume equal to 27V, ρ′ = Q/(27V) = ρ/27. Therefore the same charge is distributed through twenty-seven times as much space, making the density one twenty-seventh of its original value. Hence option C is correct.

Correct answer: B. −45 C

Explanation: The governing relation for a uniform linear charge distribution is Q = λL, where λ is linear charge density and L is the length. Substituting the given values gives Q = (−3 C/m)(15 m) = −45 C. The metre units cancel, and the negative sign remains because the distribution contains negative charge. Therefore option B is correct; option A incorrectly changes the sign, while C ignores the length factor.

Correct answer: B. −88 C

Explanation: For a uniform surface charge distribution, total charge is obtained from Q = σA, where σ is surface charge density and A is area. Using the data, Q = (−8 C/m²)(11 m²) = −88 C. The square-metre units cancel, and the negative sign indicates that the net charge is negative. Hence option B is correct. Option A has the right magnitude but wrong sign, while C uses only the area value.