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Gausss Law And Calculation Of Electric Flux

Learn gauss's law with examples, practice to calculate flux linked with any face of cube when charge is placed at center and the flux linked with a spherical surface length of a uniformly charged rod.

Electric Flux

  • Flux is defined as the number of field lines passing through surface.

\(\phi_E=\int\limits_s\vec E.d\vec a\)

  • \(\vec E.d\vec a\) is proportional to the number of lines passing through the infinitesimal area \(d\vec a\).
  • This suggests that the total flux through any closed surface depends on the total charge inside the closed surface.
  • Field lines originating from positive charge must either pass out through the surface or terminate on a negative charge inside .

  • Net flux due to charge \(q\) placed outside the closed surface is zero as numbers of field lines entering the body will be same as number of field lines leaving the body.

  • Hence, the net flux is only dependent on the charge \(q\) present inside a closed surface.

      \(\phi\propto q_{inside}\)

or, \(\phi=\dfrac {q_{inside}}{\epsilon_0}\)

where, \(\epsilon_0\) is the permittivity of the free space.

  • Thus flux is given as 

\(\phi=\oint\limits_s \vec E.d\vec a=\dfrac {Q_{inside}}{\epsilon_0}\)

 

Illustration Questions

Which among the following has higher flux linkage?

A

B

C

D Flux linkage with all is same

×

Flux linked with any surface depends on charge present inside the closed surface.

Since, charge present inside all closed surface is same.

Hence, flux linkage with all is same.

\(\phi_a=\phi_b=\phi_c\)

Which among the following has higher flux linkage?

A image
B image
C image
D

Flux linkage with all is same

Option D is Correct

Calculation of Electric Flux

  • Flux is defined as the number of field lines passing through surface \(\phi_E=\int\limits_s\vec E.d\vec a\)
  • \(\vec E.d\vec a\) is proportional to the number of lines passing through the infinitesimal area \(d\vec a\).
  • This suggests that the total flux through any closed surface depends on the total charge inside the closed surface.
  • Field lines originating from positive charge must either pass out through the surface or terminate on a negative charge inside .

  • Net flux due to charge \(q\) placed outside the closed surface is zero as number of field lines entering the body will be same as number of field lines leaving the body.

  • Hence, the net flux is only dependent on the charge q present inside a closed surface.

     \(\phi\propto q_{inside}\)

or, \(\phi=\dfrac {q_{inside}}{\mathcal E_0}\)

where, \(\mathcal E_0\) is the permittivity of the free space.

  • Thus flux is given as 

\(\phi=\oint\limits_s \vec E.d\vec a=\dfrac {Q_{inside}}{\mathcal E_0}\)

  • For any closed surface through which electric field lines are leaving, flux is always positive.
  • For any closed surface through which electric field lines are entering, flux is always negative.

Illustration Questions

Find the total flux linked with the closed surface, shown in figure.

A 0 Nm2/C

B \(\dfrac {18}{\epsilon_0}\) Nm2/C

C \(\dfrac {12}{\epsilon_0}\) Nm2/C

D \(\dfrac {6}{\epsilon_0}\) Nm2/C

×

Flux through any closed surface,

 \(\phi=\dfrac {\text {Total charge enclosed by closed surface}}{\epsilon_0}\)

\(\phi=\dfrac {8C+5C+(-1C)}{\epsilon_0}\)

     =\(\dfrac {12}{\epsilon_0}\)Nm2/C

Find the total flux linked with the closed surface, shown in figure.

image
A

0 Nm2/C

.

B

\(\dfrac {18}{\epsilon_0}\) Nm2/C

C

\(\dfrac {12}{\epsilon_0}\) Nm2/C

D

\(\dfrac {6}{\epsilon_0}\) Nm2/C

Option C is Correct

Flux due to Charge Distribution over Length of Body

  • Consider a rod with uniform charge distribution over its length L, as shown in figure.

  • Since charge is uniformly distributed over its length L.
  • So, charger per unit length of rod = \(\dfrac {Total\;charge}{Total\;length}\)
  • Consider a small Gaussian surface for the rod such that  some charge resides inside the surface.

                Then charge enclosed by closed surface is given as: -

           Charge enclosed by closed surface = Length of rod enclosed by surface × Charge per unit length

  • Charge enclosed by the closed surface will contributed to the flux.
  • Flux linked with the surface = \(\dfrac {Total\;charge\;enclosed\;by\;the\;surface}{\epsilon_0}\)

 

Illustration Questions

A rod of length L = 5 m and charge Q = +5 C which is uniformly distributed over its length is shown in figure. If a closed surface covers \(\ell=\)2.5 m of the length of rod, then calculate the amount of flux linked with this surface.

A \(\dfrac {100}{\epsilon_0}\)Nm2/C

B \(\dfrac {2.5}{\epsilon_0}\)Nm2/C

C \(2.5\epsilon_0\)Nm2/C

D \(2.5\epsilon_0^2\) Nm2/C

×

Since charge is uniformly distributed over length of rod.

So, charge per unit length = \(\dfrac {Total\; charge}{Total\; length}=\dfrac {Q}{L}\)

                  \(=\dfrac {5\,C}{5\,m}=1\) C/m

Charge enclosed by closed surface (Q') = Length enclosed by the surface × Charge per unit length

           Q' = 2.5 × 1 

           Q' = 2.5 C

Flux linked with closed surface

(\(\phi\)) = \(\dfrac {Total\; charge\; enclosed\;by\;the\;surface(Q')}{\epsilon_0}\)

                \(\phi=\dfrac {Q'}{\epsilon_0}=\dfrac {2.5}{\epsilon_0} \;Nm^2/C\)

A rod of length L = 5 m and charge Q = +5 C which is uniformly distributed over its length is shown in figure. If a closed surface covers \(\ell=\)2.5 m of the length of rod, then calculate the amount of flux linked with this surface.

image
A

\(\dfrac {100}{\epsilon_0}\)Nm2/C

.

B

\(\dfrac {2.5}{\epsilon_0}\)Nm2/C

C

\(2.5\epsilon_0\)Nm2/C

D

\(2.5\epsilon_0^2\) Nm2/C

Option B is Correct

Flux due to Uniform Charge Distribution over the Plane

  • Consider a plane with uniform charge distribution over its surface area 'A', as shown in figure.

  • Since charge is uniformly distributed over area of sheet A.

So, charge per unit Area = \(\dfrac {Total\;charge}{Total\;Area}\) = \(\dfrac {Q}{A }\)

  • Consider a small cylindrical Gaussian surface for the sheet such that some charge resides inside the surface.

          Charge enclosed by the surface = Area of sheet enclosed by Gaussian surface(A') × charge per unit area\(\left( \dfrac {Q}{A} \right)\)

  • The charge enclosed by the closed surface contributes to the flux.

Flux linked with Gaussian surface = \(\dfrac {Total\;charge\;enclosed\;by\;the\;surface}{\epsilon_0}\)

Illustration Questions

A uniformly charged square sheet of side a = 5 m is given a charge Q = +10 C. A cylindrical sheet encloses some area A'= 2 m2 of this sheet, as shown in figure. Calculate the net flux linked with this cylindrical surface.

A \({3}{\epsilon_0}\) Nm2/C

B \(\dfrac{4}{5\epsilon_0}\) Nm2/C

C \(\dfrac{4}{\epsilon_0}\) Nm2/C

D \(\dfrac{3}{\epsilon_0}\)Nm2/C

×

Since charge is uniformly distributed over area of sheet A.

So, charge per unit area = \(\dfrac {Total\;charge}{Total\;Area}\)

\(=\dfrac {10\,C}{(5)^2m^2}=\dfrac {2}{5}\,C/m^2\)

Charge enclosed by the surface (Q') = Area of sheet enclosed by surface (A') × Charge per unit Area

\(Q'=2\times\dfrac {2}{5}=\dfrac {4}{5}\,C\)

Flux linked with closed surface = \(\dfrac {Total\;charge\;enclosed\;by\;the\;surface\,(Q')}{\epsilon_0}\)

\(=\dfrac{4}{5\epsilon_0}\) Nm2/C

A uniformly charged square sheet of side a = 5 m is given a charge Q = +10 C. A cylindrical sheet encloses some area A'= 2 m2 of this sheet, as shown in figure. Calculate the net flux linked with this cylindrical surface.

image
A

\({3}{\epsilon_0}\) Nm2/C

.

B

\(\dfrac{4}{5\epsilon_0}\) Nm2/C

C

\(\dfrac{4}{\epsilon_0}\) Nm2/C

D

\(\dfrac{3}{\epsilon_0}\)Nm2/C

Option B is Correct

Flux Linked with Spherical Surface with Charges (Both Continuous & Point) Inside It

  • Consider a spherical surface with both continuous and point charge enclosed in it.
  • To calculate flux of the spherical surface, identify the point charges and continuous charge and then calculate charge on continuous charge enclosed by sphere.

Calculation of Flux on Spherical Surface

  • Consider a spherical surface which includes a part of a uniformly charged rod (continuous charge) with some point charges of different magnitude inside it, as shown in figure.
  • The total flux associated with the sphere will be 

\(\phi=\dfrac {{\text {Total charge enclosed by spherical surface}}} {\epsilon_0}\)

\(\phi=\dfrac {\text{(Charge per unit length of rod × length of rod enclosed by sphere)+all point charges inside}} {\epsilon_0}\)

 

Illustration Questions

Calculate the flux linked with a spherical surface inclosing L' = 4 m length of a uniformly charged rod Q = +5 C , L = 12 m and some point charges q1 = –8 C, q2=–3 C, q3 = 5 C, q4 = 2 C, as shown in figure.

A \(300\epsilon_0\) Nm2/C

B  \(\dfrac {11\epsilon_0}{3}\) Nm2/C

C \(110\epsilon_0\) Nm2/C

D \(\dfrac {11}{3\epsilon_0}\) Nm2/C

×

Flux linked with spherical surface   =  \(\dfrac {\text {Total charge enclosed by spherical surface}} {\epsilon_0}\)

Total charge enclosed by spherical surface

= Sum of point charges inside sphere + charge on the length L' enclosed by rod

= (–3 C) + (+5 C) + [ Charge per unit length × length of rod]

\(=2\,C+ \left( \dfrac {5}{12}\times4 \right) \)

\(=2\,+\dfrac {5}{3}=\dfrac {11}{3}C\)

Net Flux with sphere = \(\dfrac {11/3}{\epsilon_0}=\dfrac {11}{3\,\epsilon_0}\) Nm2/C

Calculate the flux linked with a spherical surface inclosing L' = 4 m length of a uniformly charged rod Q = +5 C , L = 12 m and some point charges q1 = –8 C, q2=–3 C, q3 = 5 C, q4 = 2 C, as shown in figure.

image
A

\(300\epsilon_0\) Nm2/C

.

B

 \(\dfrac {11\epsilon_0}{3}\) Nm2/C

C

\(110\epsilon_0\) Nm2/C

D

\(\dfrac {11}{3\epsilon_0}\) Nm2/C

Option D is Correct

Flux Linked with Any Face of Cube when Charge is Placed at Center

  • Consider a charge placed at the center of a cube of side 'a'.
  • Flux linked with cube = \(\dfrac {\text{Total charge enclosed by cube }(q_0)} {\epsilon_0}\)
  • Since charge is placed symmetrically with all faces of cube.

          So, net flux through any face of cube will be same and is given as \(=\dfrac {q}{6\epsilon_0}\)

Illustration Questions

Calculate the flux linked with face ABCD of the cube when charge q = q0 is placed at the center of cube.

A \(\dfrac{q}{2\epsilon_0}\)

B 0

C \(\dfrac{q}{6\epsilon_0}\)

D \(\dfrac{q}{4\epsilon_0}\)

×

Total flux linked with cube = \(\dfrac {\text{Total charge enclosed by cube}}{\epsilon_0}\)

\(=\dfrac {q_0}{\epsilon_0}\)

Since, charge is placed symmetrically.

So, flux through every face will be same.

Flux through each face = \(\dfrac {q}{6\epsilon_0}\)

Calculate the flux linked with face ABCD of the cube when charge q = q0 is placed at the center of cube.

image
A

\(\dfrac{q}{2\epsilon_0}\)

.

B

0

C

\(\dfrac{q}{6\epsilon_0}\)

D

\(\dfrac{q}{4\epsilon_0}\)

Option C is Correct

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