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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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Expert · Level 1View options
Total charge may be zero
Positive and negative parts may exist
Local density is zero at every point
Signed addition is necessary
Expert · Level 1View options
One coulomb per metre
Two coulomb per metre
Five coulomb per metre
Ten coulomb per metre
Expert · Level 1View options
Toward the removed part
Away from the removed part
Always zero
Along the axis
Expert · Level 1View options
The same direction
The opposite direction
Always zero
It cannot be determined
Expert · Level 1View options
Because the complete opposite half is absent
Because the field of every element is zero
Because the total charge is zero
Because the arc is a volume distribution
Expert · Level 1View options
Because transverse components cancel by symmetry
Because the total charge on the axis is zero
Because electric field is a scalar
Because all elements are at the centre
Expert · Level 1View options
Equal small charge elements occur at equal distances on both sides
The total charge of the rod is zero
The rod is circular
Electric field is a scalar
Expert · Level 1View options
They remain completely cancelled
They all become zero
They become unequal and may leave a residual component
They become volume components
Expert · Level 1View options
Use only the centre density over the whole area
Use only the edge density over the whole area
Add the charges of small area elements
Take the total charge as zero
Expert · Level 1View options
Spherical symmetry
Linear asymmetry
Square symmetry
No symmetry
Expert · Level 1View options
It increases more rapidly
It remains the same
It decreases
It becomes zero
Expert · Level 1View options
Because positive and negative charges act from different positions
Because the total charge is not zero
Because the density is always positive
Because electric field is a scalar
Expert · Level 1View options
Identify the distribution type, then density, symmetry, and vector direction
Use only the total charge to obtain the answer
Remove direction first and then add magnitudes
Ignore the density unit and estimate the answer
Expert · Level 1View options
Because contributions of infinitesimal elements are added in the limiting process
Because the total charge is never finite
Because direction is removed from every contribution
Because the charge density stops changing
Question 1ExpertLevel 1
If the average density of a linear distribution is zero, which conclusion is not always correct?
Correct answer: C
The governing idea is that average linear charge density represents a signed average, not the value at every location. A zero average can result when positive and negative contributions cancel, so the net charge may be zero and both signs may occur. However, the local density can vary from point to point and need not vanish everywhere. Therefore option C is the conclusion that is not always valid; zero average must not be confused with uniform zero density.
In a uniform linear distribution total charge is one hundred forty coulomb and length is twenty eight metre. If total charge doubles and length becomes seven times what is the new linear charge density?
Correct answer: B
Step 1: New charge is two hundred eighty coulomb and new length is one hundred ninety six metre. Step 2: Linear density is charge divided by length. Step 3: The exact value is ten by seven coulomb per metre so among the given options two is the closest.
At the centre of a uniformly charged complete ring, the electric field is zero. If a very small positively charged part of the ring is removed, in which direction will the total field tend?
Correct answer: A
The governing idea is superposition and cancellation by symmetry. For the complete uniformly charged ring, the field of the small removed element was exactly cancelled by the field of the rest of the ring, so the net field was zero. After removal, the remaining field equals the negative of the removed element’s field. At the centre, the field due to a positive element points away from that element; therefore the negative of this field points toward the removed part. Hence option A is correct; option B reverses the vector reasoning, while C ignores the broken symmetry.
If a small part of a uniformly negatively charged ring is removed, how will the electric-field direction at the centre compare with the corresponding positive-charge case?
Correct answer: B
The result follows from superposition and the reversal of electric-field direction for opposite charge signs. In a complete negative ring, fields cancel at the centre. Removing one negative element leaves the negative of that element’s original field as the residual field. If the same element were positive, its field would reverse direction, and the residual field would reverse as well. Thus the negative-ring case has the opposite direction to the positive-ring case. Option B is correct; A ignores charge sign, C ignores the missing part, and D is unnecessary because the geometry is specified.
Why is the electric field at the centre of a uniformly charged semicircular arc not generally taken as zero?
Correct answer: A
The governing concept is vector superposition together with geometric symmetry. In a complete uniformly charged ring, every small charge element has a diametrically opposite partner whose field at the centre cancels it. A semicircular arc contains only one half of that pairing structure, so the opposite elements needed for complete cancellation are missing. The individual element fields are not zero, and the arc carries nonzero charge. Therefore the net field is generally nonzero, making option A correct; B, C, and D contradict the physical setup.
Why is the electric field at a point on the axis of a uniformly charged complete ring directed along the axis?
Correct answer: A
The governing concept is cylindrical symmetry and vector addition. Consider two equal charge elements located symmetrically on opposite sides of the ring. At an axial point, they are equally distant, so their field magnitudes are equal. Their components perpendicular to the axis point in opposite directions and cancel pairwise. Their axial components point in the same direction and add. Repeating this pairing around the ring leaves only the axial component. Thus option A is correct; the other choices confuse charge, vector nature, or geometry.
Which symmetry condition is central to determining the electric-field direction on the perpendicular bisector of a uniformly charged rod?
Correct answer: A
The key principle is mirror symmetry of the continuous linear charge distribution. For every small element on one side of the rod’s midpoint, there is an equal element at the same distance on the other side. Their field components parallel to the rod are equal and opposite, so those components cancel. Their components perpendicular to the rod point in the same direction and add. Consequently, the net field lies along the perpendicular bisector. Option A states the required condition; the other options are irrelevant or physically false.
If the charge density of the right half of a uniform rod becomes greater than that of the left half, what may happen to the components along the rod at a point on the perpendicular bisector?
Correct answer: C
Complete cancellation of the components parallel to the rod requires equal charge contributions from corresponding elements on both sides. Increasing the right-half density breaks that mirror symmetry. The corresponding elements then produce unequal fields, so their components along the rod no longer cancel exactly. A residual component can therefore remain, with its direction depending on which side has the greater effective contribution and on the sign of charge. Option C is correct; A assumes preserved symmetry, B overstates cancellation, and D confuses component direction with distribution type.
If surface charge density on a surface decreases with distance from the centre, which method is most appropriate for finding the total charge?
Correct answer: C
The governing relation for a surface distribution is dq = sigma dA, where the surface charge density sigma may vary from point to point. Since density decreases with radial distance, one constant value cannot correctly represent the entire surface. The surface must be divided into small area elements, their individual charges sigma(r)dA evaluated, and the contributions integrated: Q = integral sigma(r)dA. Therefore option C is correct. Options A and B use only one local value, while D has no physical basis.
In a volume charge distribution, if the density depends only on the radial distance from the centre and not on direction, what type of symmetry does this indicate?
Correct answer: A
Symmetry is identified by the variables on which the charge density depends. If rho is a function only of the radial distance r, then all points on a sphere of radius r have the same density, regardless of direction. Rotating the distribution about the centre does not change it, which is precisely spherical symmetry. Option A is therefore correct. Linear or square symmetry does not describe a three-dimensional radial distribution, and option D contradicts the stated directional independence.
If the linear charge density in a one-dimensional distribution increases in proportion to the square of distance, how will the charge in equal-length sections farther from the reference point compare?
Correct answer: A
Linear charge density is charge per unit length, so the charge in a small equal-length section is approximately dq = lambda(x) dx. If lambda(x) is proportional to x squared, then sections located at larger x have larger density and therefore larger charge for the same dx. Moreover, the rate of increase itself grows with distance because a quadratic dependence is faster than a linear one. Hence option A is correct. B would require constant density, while C and D contradict the stated positive quadratic growth.
If a charge distribution has zero total charge but has dipole-like separation, why is the electric field not zero everywhere?
Correct answer: A
The governing distinction is between net charge and the spatial distribution of charge. Zero total charge means only that the algebraic sum of positive and negative charges is zero. In a dipole-like arrangement, the charges occupy different positions, so at a general observation point their distances and field directions are different. Their electric-field vectors therefore do not cancel everywhere, although they may cancel at selected points because of symmetry. Option A is correct; B contradicts the premise, and C and D misstate charge density or field nature.
What is the best order to start solving a difficult continuous charge distribution problem?
Correct answer: A
The governing method for a continuous charge distribution is to model the charge correctly before integrating its electric-field contributions. First identify whether the charge lies on a line, surface, or volume and choose the corresponding density. Next inspect symmetry to simplify the vector sum. Finally retain the direction of each contribution while integrating. Option B loses spatial information, C incorrectly discards vectors, and D can produce dimensional errors.
Why is integration called a limiting sum in continuous charge distribution?
Correct answer: A
Integration is called a limiting sum because a continuous distribution is conceptually divided into many small charge elements. For example, a line element carries dq = lambda dx, and the field is obtained by summing dE over all elements. As the element size approaches zero and the number of elements approaches infinity, the discrete sum becomes the integral. Option B is not required, while C and D misunderstand vector addition and density.
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