A point charge +q is positioned at the center of curvature of a hemisphere. Determine the electric flux passing through the hemispherical surface.

Let's put an upper half hemisphere. Now, flux passing through the entire sphere=qoSince the charge +q is symmetrically placed with respect to the upper and lower halves of the hemisphere, the flux will be equally distributed, with half of the total flux emerging from each surface.

A Uniformly charged solid non-conducting sphere of uniform volume charge density and radius R is having a concentric spherical cavity of radius r. Find out electric field intensity at following points, as shown in the figure : (1) Point A (2) Point B (3) Point C (4) Centre of the sphere

We can consider that the spherical cavity is filled with charge density and also -, thereby making net charge density zero after combining. We can consider two concentric solid spheres: One of radius R and charge density and others of radius r and charge density -.

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Gauss Law

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Gauss Law

Gauss's Law is a key concept in electromagnetism that connects the electric flux passing through a closed surface to the total charge inside it. This law is especially helpful for finding electric fields in situations with symmetrical charge distributions, like spheres, cylinders, or flat surfaces. Beyond electrostatics, Gauss's Law also plays a role in deriving the magnetic field. This leads to Gauss's Law for magnetism, which states that the total magnetic flux through any closed surface is always zero, indicating that magnetic monopoles do not exist.

1.0Electric Flux

Electric flux through a surface in an electric field shows the total number of electric field lines crossing it and is a property of the field.
The SI unit of electric flux is $N m^{2} C^{- 1}$ (Gauss) or $J m C^{1}$ . Electric flux is a scalar quantity. It can be positive, negative or zero.

$d ϕ = E d s cos θ$

$d ϕ = (E cos θ) d s$

$d ϕ = E_{n} d s$

$E_{n}$ is the component of electric field in the direction of $d s$

The electric flux over the whole area = $ϕ = \int_{S} E \cdot d s = \int_{S} E_{n} d s$

If the electric field is uniform over that area then $ϕ = E \cdot S$

Special Cases:

Case 1:If the electric field is normal to the surface, then angle of electric field E

with normal will be zero.

$ϕ = ES cos 0^{\circ} \Rightarrow ϕ = ES$

Case 2:If electric field is parallel of the surface (grazing), then angle made by $E with normal = 90°.$

If electric field is parallel of the surface (grazing), then angle made by E { \vec{E}} with normal = 90°.{\90^\circ}

$ϕ = ES cos 9 0^{\circ} \Rightarrow ϕ = 0$

Case 3: Curved surface in uniform electric field

Consider a circular surface of radius R placed in a uniform electric field, as depicted.

Flux passing through the surface $ϕ = E (π R^{2})$

Now suppose, a hemispherical surface is placed in the electric field. Flux through

hemispherical surface:

$ϕ = \int E d s cos θ = E \int d s cos θ$

$\int d s cos θ$ is projection of the spherical surface Area on base.

$\int d s cos θ = π R^{2}$

$ϕ = E (π R^{2})$

Note: If two surfaces have the same number of electric field lines passing through them, the flux will be the same, regardless of shape.

$ϕ_{1} = ϕ_{2} = ϕ_{3} = E (π R^{2})$

Case 4: Flux through a closed surface

Flux through the spherical surface

$ϕ = \int E \cdot d s = \int E d s (as E is along d s (normal))$

$ϕ = \frac{1}{4 π ε _{0}} \frac{Q}{R ^{2}} \int d S (\int d s = 4 π R^{2})$

$ϕ = (\frac{1}{4 π ε _{0}} \frac{Q}{R ^{2}}) (4 π R^{2}) \Rightarrow ϕ = \frac{Q}{ε _{0}}$

Even if the charge Q is enclosed by a different closed surface, the same number of lines of force will pass through that surface.

$ϕ = \frac{Q}{ε _{0}}$

2.0 Gauss Theorem

Gauss's law, formulated by Carl F. Gauss, relates the electric field on a closed surface to the net enclosed charge. The Gaussian surface, a hypothetical closed surface, is validated through experiments and is used to calculate electric fields for symmetric charge distributions.
The surface integral of the electric field intensity over any closed hypothetical surface (Gaussian surface) in free space equals 1/ε₀ times the total charge enclosed within the surface, where ε₀ is the permittivity of free space.
If S is the Gaussian surface and $s u m_{i = 1}^{n} q_{i}$ is the total charge enclosed by the Gaussian surface, then according to Gauss's law,

$ϕ = \oint E \cdot d s = \frac{1}{ε _{0}} \sum_{i = 1}^{n} q_{i}$

Note:

(1) Flux through the Gaussian surface is independent of its shape.

(2) The flux through a Gaussian surface depends solely on the total charge enclosed within the surface.

(3) Flux through the Gaussian surface is independent of the position of charges inside the Gaussian surface.

(4) The electric field intensity at a Gaussian surface is influenced by all the charges both inside and outside the surface.

(5) In a closed surface incoming flux is taken negative, while outgoing flux is taken positive, because $\overset{n}{^}$ is taken positive in an outward direction.

(6) A flux of zero through a Gaussian surface (Φ = 0) does not necessarily mean that the electric field (E) is zero at every point on the surface. However, if the electric field is zero at every point on the surface (E = 0), then the flux must also be zero (Φ = 0)

3.0Electric Field Due to a Spherical Shell

Electric field outside the Sphere

$ϕ = \int E \cdot d s = ϕ_{net} = \frac{q _{in}}{ε _{0}} = \frac{q}{ε _{0}}$

$\int E \cdot d s = \int E d s cos 0^{\circ} = E \int d s = E (4 π R^{2})$

$E (4 π R^{2}) = \frac{q _{in}}{ε _{0}}$

$E_{out} = \frac{q}{4 π ε _{0} r ^{2}}$

Electric field inside a sphere

$ϕ = \int E \cdot d s = ϕ_{net} = \frac{q _{in}}{ε _{0}} = 0$

$\int E \cdot d s = \int E d s cos 0^{\circ} = E \int d s = E (4 π R^{2})$

$E (4 π R^{2}) = 0$

$E_{in} = 0$

4.0Electric field due to uniformly charged solid sphere

For a solid sphere having uniformly distributed charge Q and radius R.

Electric field outside the sphere

Electric field due to uniformly charged solid sphere

$ϕ = \int E \cdot d s = ϕ_{net} = \frac{q _{in}}{ε _{0}} = \frac{Q}{ε _{0}}$

$\int E \cdot d s = \int E d s cos 0^{\circ} = E \int d s = E (4 π r^{2})$

$E (4 π r^{2}) = \frac{Q}{ε _{0}}$

$E_{out} = \frac{Q}{4 π ε _{0} r ^{2}}$

Electric field inside a solid sphere

$ϕ = \int E \cdot d s = ϕ_{net} = \frac{q _{in}}{ε _{0}} = \frac{\frac{Q}{\frac{4}{3} π R ^{3}} \times \frac{4}{3} π r ^{3}}{ε _{0}} = \frac{Q r ^{3}}{ε _{0} R ^{3}}$

$\int E \cdot d s = \int E d s cos 0^{\circ} = E \int d s = E (4 π r^{2})$

$E (4 π r^{2}) = \frac{Q r ^{3}}{ε _{0} R ^{3}}$

$E = \frac{Q r}{4 π ε _{0} R ^{3}}$

$E_{in} = \frac{K Q}{R ^{3}} r$

$E = \frac{ρ r}{3 ε _{0}} (ρ = Volume Charge Density)$

5.0Electric field due to infinite line charge

$ϕ_{3} = \int E \cdot d s = \frac{q _{in}}{ε _{0}}$

$\int E d s = \frac{λ l}{ε _{0}}$

$E \int d s = \frac{λ l}{ε _{o}}$

$E (2 π r l) = \frac{λ l}{ε _{o}}$

$E = \frac{λ}{2 π ε _{o} r} = \frac{2 kλ}{r}$

6.0Electric field due to infinitely long charged Tube

E outside the tube

$ϕ_{net} = \frac{q _{in}}{ε _{o}} = \frac{σ 2 π Rl}{ε _{o}}$

$\Rightarrow E_{out} \times 2 π r l = \frac{σ 2 π Rl}{ε _{o}}$

$E = \frac{σ R}{r ε _{o}}$

E inside the tube

$ϕ_{net} = \frac{q _{in}}{ε _{o}} = 0$

So $E_{in} = 0$

7.0Electric Field due to infinitely long solid cylinder of radius R

E at outside point:

$E \times 2 π r l = \frac{q _{in}}{ε _{o}} = \frac{ρ \times π R ^{2} l}{ε _{o}}$

$E_{out} = \frac{ρ R ^{2}}{2 r ε _{o}}$

E at inside point:

$E \times 2 π r l = \frac{q _{in}}{ε _{o}} = \frac{ρ \times π r ^{2} l}{ε _{o}}$

$\Rightarrow E_{in} = \frac{ρ r}{2 ε _{o}}$

E at inside point: Electric Field due to infinitely long solid cylinder of radius R

graph of Electric Field due to infinitely long solid cylinder of radius R

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Gauss Law

Gauss's Law is a key concept in electromagnetism that connects the electric flux passing through a closed surface to the total charge inside it. This law is especially helpful for finding electric fields in situations with symmetrical charge distributions, like spheres, cylinders, or flat surfaces. Beyond electrostatics, Gauss's Law also plays a role in deriving the magnetic field. This leads to Gauss's Law for magnetism, which states that the total magnetic flux through any closed surface is always zero, indicating that magnetic monopoles do not exist.

1.0Electric Flux

Electric flux through a surface in an electric field shows the total number of electric field lines crossing it and is a property of the field.
The SI unit of electric flux is $N m^{2} C^{- 1}$ (Gauss) or $J m C^{1}$ . Electric flux is a scalar quantity. It can be positive, negative or zero.

$d ϕ = E d s cos θ$

$d ϕ = (E cos θ) d s$

$d ϕ = E_{n} d s$

$E_{n}$ is the component of electric field in the direction of $d s$

The electric flux over the whole area = $ϕ = \int_{S} E \cdot d s = \int_{S} E_{n} d s$

If the electric field is uniform over that area then $ϕ = E \cdot S$

Special Cases:

Case 1:If the electric field is normal to the surface, then angle of electric field E

with normal will be zero.

$ϕ = ES cos 0^{\circ} \Rightarrow ϕ = ES$

Case 2:If electric field is parallel of the surface (grazing), then angle made by $E with normal = 90°.$

$ϕ = ES cos 9 0^{\circ} \Rightarrow ϕ = 0$

Case 3: Curved surface in uniform electric field

Consider a circular surface of radius R placed in a uniform electric field, as depicted.

Flux passing through the surface $ϕ = E (π R^{2})$

Now suppose, a hemispherical surface is placed in the electric field. Flux through

hemispherical surface:

$ϕ = \int E d s cos θ = E \int d s cos θ$

$\int d s cos θ$ is projection of the spherical surface Area on base.

$\int d s cos θ = π R^{2}$

$ϕ = E (π R^{2})$

Note: If two surfaces have the same number of electric field lines passing through them, the flux will be the same, regardless of shape.

$ϕ_{1} = ϕ_{2} = ϕ_{3} = E (π R^{2})$

Case 4: Flux through a closed surface

Flux through the spherical surface

$ϕ = \int E \cdot d s = \int E d s (as E is along d s (normal))$

$ϕ = \frac{1}{4 π ε _{0}} \frac{Q}{R ^{2}} \int d S (\int d s = 4 π R^{2})$

$ϕ = (\frac{1}{4 π ε _{0}} \frac{Q}{R ^{2}}) (4 π R^{2}) \Rightarrow ϕ = \frac{Q}{ε _{0}}$

Even if the charge Q is enclosed by a different closed surface, the same number of lines of force will pass through that surface.

$ϕ = \frac{Q}{ε _{0}}$

2.0 Gauss Theorem

Gauss's law, formulated by Carl F. Gauss, relates the electric field on a closed surface to the net enclosed charge. The Gaussian surface, a hypothetical closed surface, is validated through experiments and is used to calculate electric fields for symmetric charge distributions.
The surface integral of the electric field intensity over any closed hypothetical surface (Gaussian surface) in free space equals 1/ε₀ times the total charge enclosed within the surface, where ε₀ is the permittivity of free space.
If S is the Gaussian surface and $s u m_{i = 1}^{n} q_{i}$ is the total charge enclosed by the Gaussian surface, then according to Gauss's law,

$ϕ = \oint E \cdot d s = \frac{1}{ε _{0}} \sum_{i = 1}^{n} q_{i}$

Note:

(1) Flux through the Gaussian surface is independent of its shape.

(2) The flux through a Gaussian surface depends solely on the total charge enclosed within the surface.

(3) Flux through the Gaussian surface is independent of the position of charges inside the Gaussian surface.

(4) The electric field intensity at a Gaussian surface is influenced by all the charges both inside and outside the surface.

(5) In a closed surface incoming flux is taken negative, while outgoing flux is taken positive, because $\overset{n}{^}$ is taken positive in an outward direction.

(6) A flux of zero through a Gaussian surface (Φ = 0) does not necessarily mean that the electric field (E) is zero at every point on the surface. However, if the electric field is zero at every point on the surface (E = 0), then the flux must also be zero (Φ = 0)

3.0Electric Field Due to a Spherical Shell

Electric field outside the Sphere

$ϕ = \int E \cdot d s = ϕ_{net} = \frac{q _{in}}{ε _{0}} = \frac{q}{ε _{0}}$

$\int E \cdot d s = \int E d s cos 0^{\circ} = E \int d s = E (4 π R^{2})$

$E (4 π R^{2}) = \frac{q _{in}}{ε _{0}}$

$E_{out} = \frac{q}{4 π ε _{0} r ^{2}}$

Electric field inside a sphere

$ϕ = \int E \cdot d s = ϕ_{net} = \frac{q _{in}}{ε _{0}} = 0$

$\int E \cdot d s = \int E d s cos 0^{\circ} = E \int d s = E (4 π R^{2})$

$E (4 π R^{2}) = 0$

$E_{in} = 0$

4.0Electric field due to uniformly charged solid sphere

For a solid sphere having uniformly distributed charge Q and radius R.

Electric field outside the sphere

$ϕ = \int E \cdot d s = ϕ_{net} = \frac{q _{in}}{ε _{0}} = \frac{Q}{ε _{0}}$

$\int E \cdot d s = \int E d s cos 0^{\circ} = E \int d s = E (4 π r^{2})$

$E (4 π r^{2}) = \frac{Q}{ε _{0}}$

$E_{out} = \frac{Q}{4 π ε _{0} r ^{2}}$

Electric field inside a solid sphere

$ϕ = \int E \cdot d s = ϕ_{net} = \frac{q _{in}}{ε _{0}} = \frac{\frac{Q}{\frac{4}{3} π R ^{3}} \times \frac{4}{3} π r ^{3}}{ε _{0}} = \frac{Q r ^{3}}{ε _{0} R ^{3}}$

$\int E \cdot d s = \int E d s cos 0^{\circ} = E \int d s = E (4 π r^{2})$

$E (4 π r^{2}) = \frac{Q r ^{3}}{ε _{0} R ^{3}}$

$E = \frac{Q r}{4 π ε _{0} R ^{3}}$

$E_{in} = \frac{K Q}{R ^{3}} r$

$E = \frac{ρ r}{3 ε _{0}} (ρ = Volume Charge Density)$

5.0Electric field due to infinite line charge

$ϕ_{3} = \int E \cdot d s = \frac{q _{in}}{ε _{0}}$

$\int E d s = \frac{λ l}{ε _{0}}$

$E \int d s = \frac{λ l}{ε _{o}}$

$E (2 π r l) = \frac{λ l}{ε _{o}}$

$E = \frac{λ}{2 π ε _{o} r} = \frac{2 kλ}{r}$

6.0Electric field due to infinitely long charged Tube

E outside the tube

$ϕ_{net} = \frac{q _{in}}{ε _{o}} = \frac{σ 2 π Rl}{ε _{o}}$

$\Rightarrow E_{out} \times 2 π r l = \frac{σ 2 π Rl}{ε _{o}}$

$E = \frac{σ R}{r ε _{o}}$

E inside the tube

$ϕ_{net} = \frac{q _{in}}{ε _{o}} = 0$

So $E_{in} = 0$

7.0Electric Field due to infinitely long solid cylinder of radius R

E at outside point:

$E \times 2 π r l = \frac{q _{in}}{ε _{o}} = \frac{ρ \times π R ^{2} l}{ε _{o}}$

$E_{out} = \frac{ρ R ^{2}}{2 r ε _{o}}$

E at inside point:

$E \times 2 π r l = \frac{q _{in}}{ε _{o}} = \frac{ρ \times π r ^{2} l}{ε _{o}}$

$\Rightarrow E_{in} = \frac{ρ r}{2 ε _{o}}$

Frequently Asked Questions

A point charge +q is positioned at the center of curvature of a hemisphere. Determine the electric flux passing through the hemispherical surface.

A Uniformly charged solid non-conducting sphere of uniform volume charge density and radius R is having a concentric spherical cavity of radius r. Find out electric field intensity at following points, as shown in the figure : (1) Point A (2) Point B (3) Point C (4) Centre of the sphere

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Gauss Law

1.0Electric Flux

2.0 Gauss Theorem

3.0Electric Field Due to a Spherical Shell

4.0Electric field due to uniformly charged solid sphere

5.0Electric field due to infinite line charge

6.0Electric field due to infinitely long charged Tube

7.0Electric Field due to infinitely long solid cylinder of radius R

Table of Contents

Frequently Asked Questions

A point charge +q is positioned at the center of curvature of a hemisphere. Determine the electric flux passing through the hemispherical surface.

A Uniformly charged solid non-conducting sphere of uniform volume charge density and radius R is having a concentric spherical cavity of radius r. Find out electric field intensity at following points, as shown in the figure : (1) Point A (2) Point B (3) Point C (4) Centre of the sphere

Join ALLEN!

Gauss Law

1.0Electric Flux

2.0 Gauss Theorem

3.0Electric Field Due to a Spherical Shell

4.0Electric field due to uniformly charged solid sphere

5.0Electric field due to infinite line charge

6.0Electric field due to infinitely long charged Tube

7.0Electric Field due to infinitely long solid cylinder of radius R

Table of Contents