Q.Due to the presence of a point charge at the centre of a spherical Gaussian surface of diameter a, 10^6 N m^2/C amount of electric flux passes through it. Keeping the point charge at the centre, the Gaussian surface is changed to a cubical Gaussian surface of side a. The flux through the new Gaussian surface will be

Showing the 12 most recent of 37 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.S.I. unit of electric flux is (A) Vm (B) Vm^2 (C) Jm (D) NC^-1

   ›Reveal solutionSolution

   Φ = E·A → (V/m)(m²) = V·m.

   Electric flux is $\Phi = \vec{E}\cdot\vec{A}$.

   Unit of electric field $E$ = N/C = V/m (volt per metre).

   Unit of area $A$ = m².

   …

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2. CBSE 2026Set A1 markMCQ
   Q.The surface charge densities on the surface of two conducting spheres of radii r1 and r2 are equal. The ratio of electric field intensities on the surfaces is (A) r1/r2 (B) r1^2/r2^2 (C) r2^2/r1^2 (D) 1 : 1

   ›Reveal solutionSolution

   Just outside a charged conductor E = σ/ε₀; with equal σ the fields are equal (1:1).

   The electric field just outside the surface of a charged conductor is

   $E = \frac{\sigma}{\varepsilon_0},$

   which depends only on the local surface charge density σ, not on the radius.

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Electric flux is a

   (a) scalar quantity

   (b) vector quantity

   (c) scalar or vector quantity

   (d) constant quantity

   ›Reveal solutionSolution

   Electric flux is a scalar quantity, even though it is defined using two vectors.

   Electric flux through a surface is defined as

   $\Phi = \oint \vec{E}\cdot d\vec{A}$

   Although $\vec{E}$ (electric field) and $d\vec{A}$ (area vector, normal to the surface element) are both vectors, their dot product $\vec{E}\cdot d\vec{A} = E\,dA\cos\theta$ is a single number (magnitude only, with a sign depending on $\theta$) — it …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.A charge Q, is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

   (a) decrease to half

   (b) increase two times

   (c) remain unchanged

   (d) increase four times

   ›Reveal solutionSolution

   Gauss's law: flux through any closed surface = Q_enclosed / epsilon_0, and this does NOT depend on the surface's size or shape.

   Gauss's law states that for any closed (Gaussian) surface,

   flux (phi) = Q_enclosed / epsilon_0

   Here the same charge Q sits at the centre of the sphere both before and after the radius is doubled - the enclosed charge Q_enclosed is unchanged. Since flux depends ONLY on Q_enclosed and the permittivity of free space epsilon_0 (both unchanged here), the flux does not change even though the surface area (4piR^2) has increased fourfold. Doubling R spreads the same total flux over 4 ti …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.A hollow metal sphere of radius $R$ is uniformly charged. The electric field due to the sphere at a distance $r$ from the centre

   (a) increases as $r$ increases for $r<R$ and for $r>R$

   (b) is zero as $r$ increases for $r<R$ and decreases as $r$ increases for $r>R$

   (c) is zero as $r$ increases for $r<R$ and increases as $r$ increases for $r>R$

   (d) decreases as $r$ increases for $r<R$ and for $r>R$

   ›Reveal solutionSolution

   A charged conducting (hollow metal) sphere carries all its charge on the outer surface. Gauss's law gives $E=0$ inside and $E\propto 1/r^2$ (decreasing) outside.

   Setting up Gauss's law

   For a hollow, uniformly charged conducting sphere of radius $R$ and total charge $Q$, all the charge resides on the outer surface (a fundamental property of conductors in electrostatic equilibrium — free charges repel each other and move to the surface where the electric field inside the conducting material is zero).

   Take a concentric spherical Gaussian surface of radius $r$.

   Case 1: $r<R$ (inside the shell)

   The Gaussian surface of radius $r$ encloses no charge, because all the charge $Q$ lies on the surface at radius $R>r$.

   $\oint \vec E\cdot d\vec A = \frac{Q_{\text{enc}}}{\varepsilon_0} = 0 \implies E=0$

   This is true for every $r<R$ — the field is zero throughout the interior, it does not "increase" or "decrease," it is simply zero.

   Case 2: $r>R$ (outside the shell)

   Now the Gaussian surface encloses the entire charge $Q$. By spherical symmetry, $\vec E$ is radial and has the same magnitude everywhere on the Gaussian sphere, so

   …

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6. CBSE 2026Set ANNUAL1 mark
   Q.Match the Column A with Column B and write the correct pair. Column A: Electric flux. Column B:

   (i) μ₀nI,

   (ii) Volt × second,

   (iii) μ₀nI/2,

   (iv) Volt × meter,

   (v) Volt × meter⁻¹,

   (vi) Volt.

   ›Reveal solutionSolution

   Electric flux φ_E = E·A; unit = (V/m)(m²) = V·m, option (iv).

   Electric flux through a surface is φ_E = E·A (for a uniform field perpendicular to area A). The SI unit of electric field E is volt per metre (V/m) = N/C, and area is in m². Therefore the unit of electric flux is

   …

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7. CBSE 2025Set D1 markMCQ
   Q.Gauss's law states that the electric flux through a closed surface is (A) proportional to the charge enclosed (B) inversely proportional to the charge enclosed (C) zero (D) proportional to the square of the charge enclosed

   ›Reveal solutionSolution

   Gauss's law states the total electric flux through a closed surface equals the enclosed charge divided by ε₀, so Φ ∝ q_enclosed.

   Gauss's law is written as

   $\oint \vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}$

   The left side is the total electric flux Φ through the closed (Gaussian) surface. Thus

   $\Phi = \frac{q_{enc}}{\varepsilon_0}$

   …

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8. CBSE 2025Set D1 markMCQ
   Q.Inside a closed surface n electric dipoles are situated. The electric flux coming out from the closed surface will be (A) q/ε0 (B) 2q/ε0 (C) nq/ε0 (D) zero

   ›Reveal solutionSolution

   A dipole has zero net charge, so n dipoles enclose zero charge and the net flux is zero.

   Gauss's law states the net electric flux out of a closed surface is Φ = q_enclosed/ε₀.

   Each electric dipole consists of +q and −q; its net charge is +q + (−q) = 0. With n dipoles inside, the total enclosed charge is n × 0 = 0.

   …

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9. CBSE 2025Set A1 mark
   Q.Write True or False: Inside a conductor, electrostatic field is zero.

   ›Reveal solutionSolution

   The statement is True: the electrostatic field inside a conductor is zero in equilibrium.

   In electrostatic equilibrium, free charges in a conductor redistribute themselves on the surface such that the electric field inside the body of the conductor is exactly zero. If there were a residual field inside, it would exert a force on the free electrons, causing them to keep moving — contradicting the assumption …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The SI unit of electric flux is

    (a) N C m^2

    (b) N C^-1 m^-2

    (c) N C^-1 m^2

    (d) N^2 C^-1 m^2

    ›Reveal solutionSolution

    Electric flux is the field strength times the perpendicular area through which it passes, so its unit is simply the product of the units of $E$ and area.

    Electric flux through a surface is defined as $\Phi_E = \vec{E}\cdot\vec{A}$ (or $\int \vec{E}\cdot d\vec{A}$ for a general surface).

    • SI unit of electric field $E$ = N C$^{-1}$
    • SI unit of area $A$ = m$^2$ …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.The SI unit of electric flux is:

    (a) $Nm^2C^{-2}$

    (b) $NC^{-1}m^2$

    (c) $CN^2m^{-1}$

    (d) $C^2N^{-1}m^{-2}$

    ›Reveal solutionSolution

    Electric flux $\Phi_E = \vec{E}\cdot\vec{A}$, so its unit is the unit of $E$ (N C⁻¹) times the unit of area (m²).

    Electric flux through a surface is defined as $\Phi_E = \oint \vec{E}\cdot d\vec{A}$, i.e. the product of the electric field and the area component perpendicular to it.

    Since the SI unit of electric field $E$ is newton per coulomb ($NC^{-1}$) and the unit of area $A$ is square metre ($m^2$), the unit of electric flux is:

    $NC^{-1} \times m^2 = NC^{-1}m^2$

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.The SI unit of the surface integral of electric field is –

    (a) Vm

    (b) V

    (c) $NC^{-1}$

    (d) $Cm^{-3}$

    ›Reveal solutionSolution

    Electric flux $\Phi_E=\oint\vec{E}\cdot d\vec{A}$ has SI unit V·m (equivalent to N·m$^2$/C).

    The surface integral of the electric field, $\Phi_E = \oint \vec{E}\cdot d\vec{A}$, is the electric flux. Since $E$ has SI unit V/m (or equivalently N/C) and area has unit m$^2$, the flux has uni …

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