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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.
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Medium · Level 6View options
As if the whole charge were at the centre
As if there were no charge
As if charge were only at one surface point
As if field were zero everywhere
Medium · Level 6View options
It becomes double
It becomes half
It becomes four times
It remains same
Medium · Level 6View options
It becomes half
It becomes double
It remains same
It decreases four times
Medium · Level 6View options
Along the surface
Perpendicular to the surface
Circular
In any random direction
Medium · Level 6View options
Positive
Negative
Zero
Depends on distance between charges
Medium · Level 6View options
Sum of both charges
Only the positive charge inside
Only the negative charge outside
Only by shape of surface
Medium · Level 6View options
Because electric field is not uniform or simply directed on the surface
Because Gauss's law becomes false
Because enclosed charge becomes zero
Because flux has no unit
Medium · Level 6View options
Because it is an imaginary closed surface for calculation
Because electric field never exists on real surfaces
Because only metal surfaces are valid
Because surface area must be zero
Medium · Level 6View options
Negative
Zero
Positive
Cannot be determined
Medium · Level 6View options
Positive
Negative
Zero
Always infinite
Medium · Level 6View options
It remains same
It becomes half
It becomes double
It becomes zero
Medium · Level 6View options
It doubles
It becomes half
It remains same
It decreases four times
Medium · Level 6View options
One fourth
Half
Same
Double
Medium · Level 6View options
Because enclosed charge is same but area increases
Because enclosed charge increases
Because electric field is scalar
Because Gauss's law does not apply
Medium · Level 6View options
Along the wire
Radially outward
Radially inward
Circular
Medium · Level 6View options
Because field is parallel to the flat ends
Because the ends have zero area
Because the wire creates no field
Because the Gaussian surface is open
Medium · Level 6View options
Because field is parallel to the curved part
Because curved part has zero area
Because field passes only through curved part
Because sheet has no charge
Medium · Level 6View options
Where surface charge density is higher
Where surface colour is lighter
Where conductor is always thicker
Where there is no charge
Medium · Level 6View options
Because it would move free charges along the surface
Because conductors have no charges
Because field is always scalar
Because surface area is zero
Medium · Level 6View options
When surface charge density is uniform
When sheet is small and edges dominate
When charge is only at one corner
When sheet has no symmetry
Medium · Level 6View options
It decreases
It remains same
It increases
It is always zero
Medium · Level 6View options
Maximum
Zero
Infinite
Equal to surface field
Medium · Level 6View options
Total charge placed at the centre
Only half surface charge
Zero everywhere
Field of a wire
Medium · Level 6View options
Product of electric field and total area
Product of electric field and mass
Only area
Always zero
Medium · Level 6View options
Maximum
Zero
Always negative
Always positive
Question 1MediumLevel 6
How is the field outside a uniformly charged spherical shell treated?
Correct answer: A
Step 1: A spherical shell has spherical symmetry. Step 2: For an outside point, the Gaussian surface encloses the whole charge. Step 3: The result is like a point charge placed at the centre.
For an infinite line charge, how does electric field change when distance is doubled?
Correct answer: B
Step 1: Electric field of an infinite line charge is inversely proportional to distance. Step 2: When distance is doubled, the field becomes half. Step 3: For line charge, remember inverse relation with distance.
For an infinite charged plane sheet, what happens to electric field when distance is doubled?
Correct answer: C
Step 1: Electric field of an infinite plane sheet does not depend on distance. Step 2: Even when distance is doubled, field remains the same. Step 3: Remember the difference between distance rules for plane sheet and point charge.
For an infinite plane sheet with uniform surface charge density, what is the direction of electric field?
Correct answer: B
Step 1: An infinite plane sheet has symmetry along all directions in the plane. Step 2: Therefore field cannot prefer any direction along the sheet and must be normal to it. Step 3: Remember normal direction for plane sheet.
Two equal and opposite charges are inside a closed surface. According to Gauss's law, what is the net flux?
Correct answer: C
Step 1: Net flux through a closed surface depends on net enclosed charge. Step 2: Equal positive and negative charges add to zero. Step 3: Therefore net flux is zero, even though electric field may exist inside.
If a positive charge is inside a closed surface and a negative charge is placed outside it, what determines net flux?
Correct answer: B
Step 1: In Gauss's law, only charge enclosed by the closed surface is counted. Step 2: The outside negative charge may change local field, but not net flux. Step 3: If positive charge is inside, net flux is determined by it.
Gauss's law can give net flux for an asymmetric charge distribution, but why may it not directly give electric field?
Correct answer: A
Step 1: Net flux is a total over the whole closed surface. Step 2: To find electric field, the field must be separated in a simple way. Step 3: In asymmetry this is not possible, so direct field calculation is difficult.
Why need not a Gaussian surface be a real physical surface?
Correct answer: A
Step 1: The purpose of a Gaussian surface is to simplify calculation. Step 2: It may be an imaginary closed surface and need not be a real object. Step 3: Choose it according to symmetry, not material.
More field lines leave a closed surface than enter it. According to Gauss's law, what is the net charge inside?
Correct answer: C
Step 1: Outgoing field lines give positive flux. Step 2: If more lines leave than enter, net flux is positive. Step 3: Positive net flux means positive net charge inside.
More field lines enter a closed surface than leave it. What is the sign of net charge inside?
Correct answer: B
Step 1: For a closed surface, outward direction is taken positive. Step 2: Entering lines give negative flux. Step 3: If more lines enter, net flux is negative and enclosed net charge is negative.
If a positive charge at the centre of a spherical Gaussian surface is doubled, what happens to the electric field on the surface?
Correct answer: C
Step 1: Radius of the spherical surface is unchanged, so distance is unchanged. Step 2: Field of a point charge is proportional to source charge. Step 3: If charge doubles, field on the surface doubles.
For a point charge at the centre, what happens to total flux if the radius of spherical Gaussian surface is doubled?
Correct answer: C
Step 1: Total flux depends on net charge enclosed by the closed surface. Step 2: Changing radius does not change enclosed charge. Step 3: Therefore total flux remains same, though field value may change.
For a point charge at the centre, what happens to electric field on the spherical Gaussian surface if its radius is doubled?
Correct answer: A
Step 1: Electric field of a point charge varies inversely as square of distance. Step 2: When radius doubles, square of distance becomes four times. Step 3: Hence field on the surface becomes one fourth.
For a spherical Gaussian surface with a point charge at centre, why does total flux remain same while field decreases when radius increases?
Correct answer: A
Step 1: Total flux is decided only by enclosed charge, so it stays same. Step 2: On a larger sphere, the same flux spreads over larger area. Step 3: Therefore electric field per unit area decreases.
A long charged wire is positive. What is the direction of electric field on the curved part of a cylindrical Gaussian surface?
Correct answer: B
Step 1: Field from a positive line charge points outward. Step 2: On the curved surface of the cylinder, the field is radial and normal to the surface. Step 3: For a positive wire, remember radially outward direction.
For a long charged wire, why is flux through the flat ends of a cylindrical Gaussian surface zero?
Correct answer: A
Step 1: Field of a long wire is radial. Step 2: For the flat ends of the cylinder, the field lies parallel to those surfaces. Step 3: A field parallel to a surface gives zero flux through it.
For an infinite plane sheet, why is flux through the curved part of the cylindrical Gaussian surface zero?
Correct answer: A
Step 1: Field of an infinite plane sheet is perpendicular to the sheet. Step 2: For a pillbox Gaussian surface, this field is parallel to the curved side. Step 3: Therefore flux through the curved side is zero.
Where will electric field just outside a conductor be greater?
Correct answer: A
Step 1: Field just outside a conductor is related to surface charge density. Step 2: Higher surface charge density gives stronger electric field. Step 3: Remember stronger field near sharper regions.
Why can there be no tangential component of electric field on a conductor surface in electrostatic equilibrium?
Correct answer: A
Step 1: Free charges in a conductor can move easily. Step 2: If a field component existed along the surface, charges would keep moving. Step 3: Therefore, in electrostatic equilibrium, field is normal to the surface.
In which situation does Gauss's law give distance-independent field for an infinite plane sheet?
Correct answer: A
Step 1: An infinite uniformly charged sheet has strong symmetry. Step 2: This symmetry gives equal fields on both sides independent of distance. Step 3: For finite sheets, edge effects can disturb this simple result.
Inside a solid non-conducting sphere with uniform volume charge density, how does electric field generally change with distance from centre?
Correct answer: C
Step 1: Inside a uniformly charged solid non-conducting sphere, enclosed charge increases with radius. Step 2: As a result, field generally increases as we move outward from the centre. Step 3: Remember the difference between conductor and non-conducting solid sphere.
What is the electric field at the centre of a uniformly charged solid non-conducting sphere?
Correct answer: B
Step 1: At the centre of the sphere, symmetry exists in all directions. Step 2: Effects from opposite directions balance each other. Step 3: Therefore, electric field at the centre is zero.
Outside a uniformly charged solid non-conducting sphere, the electric field is like that of what?
Correct answer: A
Step 1: Outside the sphere, a Gaussian surface encloses the whole charge. Step 2: Due to spherical symmetry, outside field is like total charge at the centre. Step 3: From outside, a spherical charge distribution behaves like a point charge.
On a Gaussian surface, electric field has same magnitude everywhere and is normal to the surface. How is total flux found?
Correct answer: A
Step 1: For uniform and normal field, each small part contributes simply. Step 2: Total flux is electric field multiplied by total area. Step 3: This is the simplicity of symmetry-based Gauss calculation.
If electric field is parallel to a part of a Gaussian surface, what will be the flux through that part?
Correct answer: B
Step 1: Flux is related to field crossing a surface. Step 2: A field parallel to the surface does not pass through that part. Step 3: Therefore flux through that part is zero.
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