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In Class 12 Physics, this topic introduces continuous charge distribution, where electric charge is spread smoothly along a line, over a surface, or throughout a volume rather than concentrated at separate points. Students learn linear, surface, and volume charge densities and use small charge elements with integration to calculate total charge and electric fields. The topic strengthens their understanding of superposition and prepares them to analyse charged rods, rings, discs, sheets, and other extended systems in the chapter Electric Charges and Fields.
TOPIC PRACTICE
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Up to 25 questions from this page. Select your focus, then start.
25 questions
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Easy · Level 11View options
The longest part
The shortest part
All parts have equal charge
No part has charge
Easy · Level 11View options
The larger-area part
The smaller-area part
All parts have equal charge
No part has charge
Easy · Level 11View options
The larger-volume part
The smaller-volume part
All parts have equal charge
No part has charge
Easy · Level 11View options
A small length
A small area
A small volume
A small time interval
Easy · Level 11View options
A small length
A small area
A small volume
A small distance
Easy · Level 11View options
A small length
A small area
A small volume
A small force
Easy · Level 11View options
It remains the same
It doubles
It becomes half
It becomes four times
Easy · Level 11View options
It becomes three times
It becomes four times
It becomes one fourth
It remains the same
Easy · Level 11View options
Three sevenths
Four sevenths
One seventh
One half
Easy · Level 11View options
Seven twelfths
Five twelfths
Five sixths
One twelfth
Easy · Level 11View options
One tenth
One half
Nine tenths
Whole
Easy · Level 11View options
Longest part
Shortest part
Middle-length part
All are the same
Easy · Level 11View options
Largest-area part
Smallest-area part
All are the same
Density becomes zero
Easy · Level 11View options
Largest-volume part
Smallest-volume part
All are the same
Density becomes infinite
Easy · Level 11View options
Positive
Negative
Zero
Signless
Easy · Level 11View options
Positive
Negative
Zero
Always signless
Easy · Level 11View options
The density will be zero
The density will be positive
The density will be negative
The density will be infinite
Easy · Level 11View options
Zero
Positive
Negative
Infinite
Easy · Level 11View options
+24 C
−24 C
−6 C
0 C
Easy · Level 11View options
+14 C
−14 C
−7 C
0 C
Easy · Level 11View options
+5 C/m
−5 C/m
−60 C/m
0 C/m
Easy · Level 11View options
Nine times
Three times
One-ninth
Unchanged
Easy · Level 11View options
Twenty-seven times
Nine times
One twenty-seventh
Three times
Easy · Level 11View options
+45 C
−45 C
−3 C
0 C
Easy · Level 11View options
+88 C
−88 C
−11 C
0 C
Question 1EasyLevel 11
A uniformly linearly charged rod is divided into three parts of unequal lengths. Which part has the greatest total charge?
Correct answer: A
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.
A plate has uniform surface charge density and is divided into parts of unequal areas. Which part has the greater total charge?
Correct answer: A
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.
A solid has uniform volume charge density and is divided into parts of unequal volumes. Which part has the greatest total charge?
Correct answer: A
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.
If charge density is given per square metre, what must be multiplied by it to obtain the charge of a small element?
Correct answer: B
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.
If charge density is given per cubic metre, what must be multiplied by it to obtain the charge of a small element?
Correct answer: C
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.
If charge density is given per metre, what must be multiplied by it to obtain the charge of a small element?
Correct answer: A
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.
If surface density doubles and area becomes half in a uniform surface distribution, what happens to the total charge?
Correct answer: A
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.
If volume density becomes one fourth and volume becomes twelve times, what happens to the total charge?
Correct answer: A
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.
In a uniform linear distribution, a part has length equal to three sevenths of the total length. What fraction of the total charge is on that part?
Correct answer: A
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.
In a uniform surface distribution, a part has area equal to five twelfths of the total area. What fraction of the total charge is on that part?
Correct answer: B
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.
In a uniform volume distribution, a part has volume equal to nine-tenths of the total volume. What fraction of the total charge is in that part?
Correct answer: C
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.
If a wire with uniform linear density is cut into three unequal length parts, which part has greater linear density?
Correct answer: D
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.
A uniformly surface-charged plate is divided into many parts with unequal areas. Which part has greater surface charge density?
Correct answer: C
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.
If a solid with uniform volume charge density is divided into four parts of unequal volumes, which part has greater volume charge density?
Correct answer: C
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.
In a uniform surface distribution, surface density is negative and area is positive. What is the sign of total charge?
Correct answer: B
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.
In a uniform linear distribution, total charge is negative and length is positive. What is the sign of linear charge density?
Correct answer: B
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.
On a uniform surface, the total charge is zero and the area is non-zero. What conclusion follows about the surface charge density?
Correct answer: A
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.
If linear charge density is zero in a uniform linear distribution and the length is non-zero, what is the total charge?
Correct answer: A
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.
If the volume charge density is −6 C/m³ and the volume is 4 m³, what is the total charge?
Correct answer: B
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.
In a uniform surface charge distribution, the surface charge density is −7 C/m² and the area is 2 m². What is the total charge?
Correct answer: B
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.
A wire has total charge −60 C and length 12 m. What is its uniform linear charge density?
Correct answer: B
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.
If the area is made nine times while the total charge remains unchanged in a uniform surface distribution, what happens to the surface charge density?
Correct answer: C
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.
If the volume is made twenty-seven times while the total charge remains unchanged in a uniform volume distribution, what happens to the volume charge density?
Correct answer: C
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.
If the uniform linear charge density is −3 C/m and the length is 15 m, what is the total charge?
Correct answer: B
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.
In a uniform surface charge distribution, the surface charge density is −8 C/m² and the area is 11 m². What is the total charge?
Correct answer: B
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.
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