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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.
Practice questions
01 In a diagram, the number of electric field lines leaving a closed surface is greater than the number entering it. What does this indicate about the net charge inside?
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Answer and explanation
Correct answer: A. Net charge is positive
Explanation: Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. Field lines leaving the surface contribute outward flux, while entering lines contribute inward flux. If more lines leave than enter, the net flux is positive, so Q_enclosed is positive. This is a qualitative line-density interpretation; option C would require equal net flux, not excess outward flux.
02 If the same number of electric field lines enter and leave a closed surface, what is the most suitable conclusion about the net charge?
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Answer and explanation
Correct answer: A. Net charge is zero
Explanation: Gauss’s law connects net electric flux with enclosed charge: Φ_net = Q_enclosed/ε₀. Equal numbers of entering and leaving field lines represent equal inward and outward contributions, giving zero net flux. Therefore Q_enclosed is zero. This does not imply that the electric field vanishes at every point: equal positive and negative charges may produce nonzero local fields while their total enclosed charge is zero. Hence D is too strong.
03 If the same number of field lines enter and leave a closed surface, what is the most suitable conclusion about the net charge?
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Answer and explanation
Correct answer: A. The net charge inside may be zero
Explanation: Gauss’s law connects net flux through a closed surface with enclosed net charge: Φ = Qinside/ε₀. Equal inward and outward field-line contributions give zero net flux, so the algebraic net charge inside is zero. This does not require the region to be empty; equal positive and negative charges can cancel in total. Therefore A is the suitable conclusion. It does not imply only positive or only negative charge, nor does it make the external field zero.
04 If equal magnitude positive and negative charges are inside a closed surface, must the electric field outside the surface be zero?
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Answer and explanation
Correct answer: A. No, outside field can exist even if net charge is zero
Explanation: Equal positive and negative charges give a net enclosed charge of zero, so Gauss’s law gives zero net electric flux through the closed surface. However, zero total flux does not mean the electric field is zero at every point outside. The charge arrangement may form a dipole or another nonsymmetric distribution whose external field is nonzero. Option B incorrectly confuses zero net charge or flux with zero field; option D also contradicts charge conservation in the stated arrangement.
05 If net charge inside a closed surface is positive but field is inward through some parts of the surface, what is the sign of total flux?
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Answer and explanation
Correct answer: A. Positive
Explanation: The sign of local flux depends on the direction of the electric field relative to the outward area vector, so inward field over some parts can give negative local contributions. Nevertheless, Gauss’s law determines the complete closed-surface flux: Φ = Q_enclosed/ε₀. Since the net enclosed charge is positive, the algebraic total flux is positive. The inward portions do not change this conclusion; therefore option A is correct.
06 If total flux through a closed surface is zero, which conclusion is safest?
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Answer and explanation
Correct answer: A. Net enclosed charge is zero
Explanation: Gauss’s law gives the exact relation Phi_closed = Q_enclosed/epsilon_0. Therefore, if the total flux through a closed surface is zero, the algebraic net charge enclosed by that surface must be zero. This does not mean that the electric field vanishes at every point, nor does it exclude equal positive and negative charges inside. Charges outside may still affect the local field while contributing zero net flux. Thus option A is the safest conclusion.
07 There is no charge inside a closed surface, but the field on the surface is said to be outward everywhere. Why is this doubtful?
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Answer and explanation
Correct answer: A. Because outward field everywhere would give positive net flux
Explanation: Gauss’s law states that the net electric flux through a closed surface is Φ = Q_enclosed/ε₀. With no charge enclosed, the net flux must be zero. If the electric field points outward everywhere and is nonzero, its dot product with the outward area vector is positive on every element, so the total flux would be positive rather than zero. Therefore the stated situation is doubtful, making option A correct. External charges can produce fields, but their total flux through an empty closed surface still sums to zero.
08 A closed surface has zero net charge inside, but contains a dipole. What does correct use of Gauss law tell?
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Answer and explanation
Correct answer: A. Net flux is zero but field need not be zero everywhere
Explanation: Gauss’s law states that the net electric flux through any closed surface is Φ = Q_enclosed/ε₀. A dipole contains equal positive and negative charges, so its total enclosed charge is zero and the net flux is therefore zero. However, electric flux is a surface integral, and positive and negative contributions can cancel while the electric field remains non-zero at individual points. Thus zero flux does not mean zero field everywhere.
09 Gauss law gives total flux. Why can the field at every point on an asymmetric surface not be found directly from it?
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Answer and explanation
Correct answer: A. Because field is not uniform and simply directed on the surface
Explanation: Gauss’s law determines the integral quantity ∮E·dA = Q_enclosed/ε₀, not the value of E separately at every point. In highly symmetric situations, the field magnitude and its angle with the surface are constant or known, so E can be taken outside the integral. An asymmetric surface generally lacks this property; the field varies in magnitude and direction, so one total flux equation cannot determine its pointwise values. Therefore option A is correct.
10 A closed surface has zero net charge inside, but positive and negative charges are at different positions. Which statement about total flux and field is correct?
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Answer and explanation
Correct answer: A. Total flux is zero but field need not be zero everywhere
Explanation: Gauss’s law states that the net electric flux through a closed surface is Φ = Q enclosed/ε₀. Because the enclosed positive and negative charges have algebraic sum zero, the total flux is zero. This law does not say that the electric field is zero at every point on the surface. Charges at different positions generally produce nonuniform fields whose outward and inward flux contributions cancel only in total. Thus option A is correct.
11 A positive charge is inside a closed surface. If a very large negative charge is placed outside the same surface, what happens to total flux?
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Answer and explanation
Correct answer: A. It remains due to the inside positive charge
Explanation: Gauss’s law states that the net electric flux through a closed surface is Φ = Qenclosed/ε₀. An external charge can alter the electric field at different points of the surface, but its field contributes zero net flux because it enters and leaves the closed surface in equal overall measure. Thus the total flux remains Qinside/ε₀ and is fixed by the enclosed positive charge. The outside charge does not make the flux negative or zero.
12 Total flux through a closed surface enclosing a dipole is zero, but field lines appear to cross the surface. How is this possible?
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Answer and explanation
Correct answer: A. As many lines leave as enter
Explanation: A dipole consists of equal positive and negative charges, so its net enclosed charge is zero. Gauss’s law therefore gives the algebraic total flux as Φ = Qnet/ε0 = 0. This does not mean the electric field is absent. Field lines can leave the closed surface through some regions and enter it through others; outward flux and inward flux have opposite signs and cancel in the total. Thus option A correctly explains the zero net flux.
13 The net flux through a closed surface is zero, yet an electric field exists at different parts of the surface. What is the correct conclusion?
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Answer and explanation
Correct answer: A. The net enclosed charge may be zero.
Explanation: Gauss’s law relates the net flux through a closed surface to the net enclosed charge, not to the electric field at every individual point. Positive and negative local flux contributions can cancel in the surface integral, giving zero total flux even though the field is nonzero in many regions. Thus the enclosed algebraic charge may be zero. Option B confuses zero net flux with zero field, so option A is correct.
14 If a closed surface is changed from a sphere to an irregular surface while the enclosed net charge remains the same, why does the total flux remain the same?
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Answer and explanation
Correct answer: A. Because total flux depends on the net enclosed charge.
Explanation: Gauss’s law states that the total electric flux through any closed surface is Φ = Q_enclosed/ε₀. It does not require the surface to be spherical or regular. Changing the shape can alter the field magnitude, direction, and local flux contribution at different portions, but the closed-surface integral is fixed by the same enclosed net charge. Hence option A is correct; option B is unnecessarily strong and generally false.
15 Gauss law gives total flux. Why is it difficult to directly find field at every point on an asymmetric surface?
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Answer and explanation
Correct answer: A. Because field is not uniform or simply directed on the surface
Explanation: Gauss’s law provides the integral relation ∮E·dA = Q_enclosed/ε₀, which gives the total flux rather than the electric field value at each point. In highly symmetric situations, E has the same magnitude or a simple direction over parts of the surface, allowing E to be taken outside the integral. On an asymmetric surface, both magnitude and direction generally vary from point to point, so the total flux alone cannot determine the local field.
16 Net charge inside a closed surface is negative and many positive charges are outside. What will be the sign of net flux?
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Answer and explanation
Correct answer: A. Negative
Explanation: Gauss’s law states that the net electric flux through a closed surface is Φ = Qenclosed/ε₀. Only the algebraic sum of charge enclosed by the surface determines the net flux. Charges outside may alter the electric field at individual points on the surface, but their total flux contribution through a closed surface is zero. Since the enclosed charge is negative, the net flux is negative.
17 Total flux through a closed surface is zero. Which statement is not necessarily true?
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Answer and explanation
Correct answer: A. Electric field is zero everywhere on the surface
Explanation: Gauss’s law gives Φ_total = Q_enclosed/ε₀, so zero total flux implies zero net enclosed charge. Equal positive and negative charges may produce this result, and external charges can create fields on the surface while contributing zero net flux through the closed surface. However, zero net flux does not require the field to vanish at every point; positive and negative local contributions can cancel. Thus A is not necessary.
18 A point charge is inside a spherical Gaussian surface but not at the centre. Which statement about total flux is correct?
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Answer and explanation
Correct answer: A. Total flux remains the same, but field on the surface will not be uniform
Explanation: Step 1: In Gauss's law, total flux is decided by net enclosed charge. Step 2: Whether the charge is at the centre or off-centre, if it is inside the surface the total flux is unchanged. Step 3: But field magnitude will not be same everywhere on the surface, so separate total flux from local field.
19 A closed surface contains no charge, but an external positive charge creates electric field on the surface. What is the total flux?
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Answer and explanation
Correct answer: A. Zero
Explanation: Step 1: An external charge can produce electric field on the surface. Step 2: Its lines enter and leave the closed surface equally. Step 3: Since net enclosed charge is zero, total closed flux is zero.
20 A charge is placed at the centre of a cube. By symmetry, flux through one face is what fraction of total flux?
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Answer and explanation
Correct answer: A. One sixth of the total flux
Explanation: Step 1: The charge is at the centre of the cube, so all six faces are equivalent. Step 2: Total flux is shared equally by the six faces. Step 3: Therefore flux through one face is one sixth of the total flux.
21 If a charge is placed at one corner of a cube and eight identical cubes can be joined so that the charge becomes the centre of a larger cube, what fraction of total flux belongs to the original cube?
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Answer and explanation
Correct answer: A. One eighth of the total flux
Explanation: Step 1: Joining eight identical cubes makes the corner charge the centre of a larger cube. Step 2: By symmetry, the total flux of the larger cube is shared equally by eight small cubes. Step 3: So the original cube corresponds to one eighth of total flux.
22 On a Gaussian surface, electric field magnitude is same everywhere, but it is outward on half the surface and inward on the other half. If the two areas are equal, what is the total flux?
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Answer and explanation
Correct answer: A. Zero
Explanation: Step 1: Outward field gives positive flux. Step 2: Inward field gives negative flux. Step 3: With equal magnitude and equal areas, the two contributions cancel, so total flux is zero.
23 Total flux through a closed surface is zero. Is it correct to say that there is no charge inside the surface?
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Answer and explanation
Correct answer: A. No, equal positive and negative charges may be inside
Explanation: Step 1: Zero total flux means net enclosed charge is zero. Step 2: Equal opposite charges may still be present inside. Step 3: Therefore do not conclude that no charge exists inside from zero total flux.
24 A Gaussian surface is closed and has net positive charge inside. Still field enters through some parts of the surface. What will be the sign of total flux?
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Answer and explanation
Correct answer: A. Positive
Explanation: Step 1: Field entering through some parts can give local negative flux. Step 2: Total flux is the algebraic sum over the complete closed surface. Step 3: Since net enclosed charge is positive, total flux is positive.
25 A closed surface has net negative charge inside, but field leaves through a small part of the surface. Which statement about total flux is correct?
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Answer and explanation
Correct answer: A. Total flux will be negative
Explanation: Step 1: Field leaving through a small part can give local positive flux. Step 2: But total closed flux is determined by net enclosed charge. Step 3: Since net charge is negative, total flux remains negative.
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