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In this Class 12 Physics topic from Chapter 1, Electric Charges and Fields, students learn how electric flux is related to the net charge enclosed by a closed surface through Gauss’s law. The topic develops the idea of Gaussian surfaces, uses symmetry to simplify electric-field calculations, and applies the law to charged spherical shells, uniformly charged spheres, infinite line charges, and plane sheets. It also helps students understand the electric field inside conductors and choose suitable surfaces for solving electrostatic problems.
TOPIC PRACTICE
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Up to 25 questions from this page. Select your focus, then start.
25 questions
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Expert · Level 5View options
Total flux will be zero
Total flux will be positive
Total flux will be negative
Gauss's law will not apply
Expert · Level 5View options
Total flux remains unchanged
Total flux becomes double
Total flux becomes zero
Total flux becomes negative
Expert · Level 5View options
Total flux remains same but field on the surface is not uniform
Total flux becomes zero
Total flux exists only when charge is at the centre
The spherical surface is no longer closed
Expert · Level 5View options
Unchanged
One-sixteenth
Four times
Sixteen times
Expert · Level 5View options
One-sixteenth of the earlier value
One-fourth of the earlier value
Four times the earlier value
Unchanged
Expert · Level 5View options
Because field decreases and enclosed charge remains the same
Because field never decreases
Because the sphere becomes an open surface
Because enclosed charge automatically doubles
Expert · Level 5View options
Positive
Negative
Zero
Depends on surface shape
Expert · Level 5View options
Negative
Positive
Zero
Infinite
Expert · Level 5View options
Zero
Positive
Negative
Always maximum
Expert · Level 5View options
Zero
Very large positive
Very large negative
Depends on distance
Expert · Level 5View options
Field is not simple or uniform on the chosen surface
Gauss's law has become wrong
Field cannot be found because the surface is closed
Enclosed charge must be zero
Expert · Level 5View options
Because field magnitude and direction on the surface are not simple
Because the law becomes false in asymmetric cases
Because making a closed surface is impossible
Because charges do not create field
Expert · Level 5View options
Because electric field is perpendicular to the area vector of the ends
Because flat ends have zero area
Because line charge creates no field
Because the cylinder is not a closed surface
Expert · Level 5View options
Because all points on the curved surface are at the same distance from the line
Because electric field is parallel to the side surface
Because side surface has zero area
Because the sheet creates no field
Because the pillbox surface is not closed
Expert · Level 5View options
Because field is perpendicular to the sheet on both sides
Because field passes through only one face
Because only side surface cuts the field
Because field does not exist on one side of the sheet
Expert · Level 5View options
Unchanged
One-tenth of the earlier value
One-hundredth of the earlier value
Ten times
Expert · Level 5View options
Three times
One-third
Nine times
Unchanged
Expert · Level 5View options
Symmetry explains cancellation of field at every interior point
Symmetry makes shell charge disappear
Symmetry creates charge inside
Symmetry makes Gauss's law false
Expert · Level 5View options
Because an external Gaussian sphere encloses all charge and symmetry is spherical
Because field is zero inside, so it must be zero outside
Because the shell is like a plane sheet
Because charge is actually at the centre
Expert · Level 5View options
It increases directly with distance
It decreases inversely with square of distance
It remains independent of distance
It is zero everywhere
Expert · Level 5View options
Because inside field is proportional to distance and distance at centre is zero
Because all charge gathers at the centre
Because Gauss's law does not apply at the centre
Because field is infinite at the centre
Question 1ExpertLevel 5
The net charge inside a closed surface is zero but the electric field on the surface is not zero everywhere. According to Gauss's law, which conclusion is correct?
Correct answer: A
Step 1: Gauss's law relates total flux to net enclosed charge. Step 2: If net charge is zero, total flux is zero even when local field exists. Step 3: Do not confuse zero total flux with zero field everywhere.
A charge inside a closed surface is moved from the centre to near the boundary but remains inside. What happens to total flux?
Correct answer: A
Step 1: Total flux depends on enclosed charge, not on the charge position. Step 2: The charge remains inside, so enclosed charge is unchanged. Step 3: Field distribution may change, but total flux remains the same.
A point charge is not at the centre of a Gaussian sphere but is still inside it. Which statement about total flux is correct?
Correct answer: A
Step 1: Gauss's law does not require the charge to be at the centre. Step 2: Since the charge is inside, total flux is decided by enclosed charge. Step 3: Off-centre position makes field non-uniform on the surface, so field calculation becomes harder.
The radius of a Gaussian sphere is made four times and the same point charge remains inside. What happens to total flux?
Correct answer: A
Step 1: Total flux depends on enclosed charge. Step 2: Changing radius does not change the charge inside. Step 3: Hence total flux remains unchanged, though field on the surface decreases.
For the same Gaussian sphere, if its radius is made four times, what will the electric field on the surface become?
Correct answer: A
Step 1: Field due to a point charge is inversely proportional to square of distance. Step 2: Making radius four times makes square of distance sixteen times. Step 3: So field on the surface becomes one-sixteenth.
Why can total flux remain the same even when the area of a Gaussian sphere increases?
Correct answer: A
Step 1: Field due to a point charge decreases with distance. Step 2: At the same time, area of the Gaussian sphere increases. Step 3: If enclosed charge is unchanged, the total effect gives the same flux.
A closed surface encloses five positive and three negative charges of equal magnitude. What is the sign of total outward flux?
Correct answer: A
Step 1: Total flux is decided by the sign of net enclosed charge. Step 2: Five positive and three equal negative charges give net positive charge. Step 3: Therefore total outward flux is positive.
A closed surface encloses two positive and six negative charges of equal magnitude. What will the total outward flux be?
Correct answer: A
Step 1: In Gauss's law, total flux depends on net charge inside. Step 2: Two positive and six negative charges give net negative charge. Step 3: Hence outward flux is negative.
A closed surface encloses four positive and four negative charges of equal magnitude. Field on the surface may be non-zero, yet what is the total flux?
Correct answer: A
Step 1: Four positive and four equal negative charges give zero net charge. Step 2: Individual charges may create field on the surface. Step 3: But net enclosed charge is zero, so total flux is zero.
A large positive charge is placed very close outside a Gaussian surface. If no charge is inside, what is the total flux?
Correct answer: A
Step 1: The outside charge can make the field strong on the surface. Step 2: But it is not enclosed by the surface. Step 3: Its lines enter and leave equally, so total flux remains zero.
Total flux is known from Gauss's law but electric field cannot be found directly. What is the most likely reason?
Correct answer: A
Step 1: Gauss's law gives total flux. Step 2: To find field, the field must be expressible simply on the surface. Step 3: Without symmetry, direct field calculation is difficult.
Gauss's law is valid even for an asymmetric charge distribution. Why does its use become difficult?
Correct answer: A
Step 1: Validity of the law and ease of calculation are different. Step 2: In asymmetry, field can vary from point to point. Step 3: Without symmetry, isolating field from flux becomes difficult.
For an infinite line charge, why is the flux through the flat ends of a cylindrical Gaussian surface zero?
Correct answer: A
Step 1: Field of an infinite line charge is radial. Step 2: Area vectors of flat ends of the cylinder are along the axis. Step 3: These directions are perpendicular, so end flux is zero.
For an infinite line charge, why is the field taken as same on the curved surface of the cylinder?
Correct answer: A
Step 1: An infinite line charge has cylindrical symmetry. Step 2: Every point on a coaxial cylindrical surface is equally distant from the axis. Step 3: Hence field magnitude is taken the same there.
How does the electric field change when distance from an infinite line charge is made five times?
Correct answer: A
Step 1: Field of an infinite line charge is inversely proportional to distance. Step 2: If distance becomes five times, field becomes five times smaller. Step 3: Hence it becomes one-fifth.
How does electric field change when distance from a point charge is made five times?
Correct answer: A
Step 1: Field of a point charge varies inversely with square of distance. Step 2: Making distance five times makes square of distance twenty-five times. Step 3: So field becomes one-twenty-fifth.
What is the correct distance dependence for fields of point charge, infinite line charge, and infinite sheet?
Correct answer: A
Step 1: For point charge, spherical area grows with square of distance. Step 2: For line charge, cylindrical curved area grows with distance. Step 3: For infinite sheet, field is independent of distance.
For an infinite plane sheet, why is flux through the side surface of the pillbox zero?
Correct answer: A
Step 1: Field of an infinite sheet is perpendicular to the sheet. Step 2: For the side surface of the pillbox, this field runs along the surface. Step 3: Since it does not cross the side surface, side flux is zero.
In a pillbox surface for an infinite plane sheet, why do both flat faces contribute to flux?
Correct answer: A
Step 1: An infinite uniform sheet has the same symmetry on both sides. Step 2: The field is perpendicular on both sides. Step 3: Therefore both flat faces of the pillbox contribute to flux.
Distance from an infinite plane sheet is made ten times while surface charge density remains same. What happens to electric field?
Correct answer: A
Step 1: Field of an infinite plane sheet does not depend on distance. Step 2: Its value is connected to surface charge density. Step 3: Since density is unchanged, increasing distance does not change the field.
If surface charge density of an infinite plane sheet becomes three times, how does the electric field change?
Correct answer: A
Step 1: Field of an infinite sheet is proportional to surface charge density. Step 2: Distance effect does not appear in this ideal case. Step 3: If density becomes three times, field also becomes three times.
A Gaussian surface inside a charged spherical shell encloses no charge. What role does symmetry play in saying the field inside is zero?
Correct answer: A
Step 1: The Gaussian surface inside encloses zero charge. Step 2: Zero total flux alone does not always prove zero field everywhere. Step 3: Spherical symmetry shows that the field inside is zero at every point.
Why does the field outside a charged spherical shell behave like the same charge placed at the centre?
Correct answer: A
Step 1: A spherical shell shows spherical symmetry from outside. Step 2: An outside Gaussian sphere encloses the entire charge. Step 3: Hence outside field is like that of a point charge at the centre.
Inside a uniformly volume-charged solid sphere, how does field change with distance from the centre?
Correct answer: A
Step 1: An inner Gaussian surface encloses only the charge within it. Step 2: Enclosed charge grows as cube of distance and surface area as square. Step 3: Therefore field inside is proportional to distance.
Why is electric field zero at the centre of a uniformly volume-charged solid sphere?
Correct answer: A
Step 1: Inside a uniformly charged solid sphere, field increases with distance from centre. Step 2: At the centre, distance is zero. Step 3: Therefore electric field at the centre is zero.
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