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Calculation Of Electric Field Using Gausss Law

Learn gauss law formula for electric field and Gaussian surface sphere examples, practice electric field inside a spherical shell and a Sphere, gauss's law cylinder problems and solutions.

Flux and Gauss's Law

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

            \(\phi_E = \int\limits _S \vec{E.}\,\,\vec{da}\)

  • \(\vec{E.}\,\,\vec{da}\) is proportional to the number of lines passing through the infinitesimal area  \(\vec{da}\).
  • 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 is 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.

              \(\phi \propto q_{in}\)

       or, \(\phi = \dfrac{q_{in}}{\varepsilon _0}\)

where \(\varepsilon _0\) is the permittivity of the free space.

  • Thus, flux is given as \(\phi = \oint\limits _S \vec{E.}\,\,\vec{da} = \dfrac{Q_{in}}{\varepsilon_o}\)

Illustration Questions

Consider a spherical Gaussian surface, as shown in figure. Choose the incorrect option.

A For Gauss's law  \(\oint \vec{E.}\,\vec{da} = \dfrac {\Sigma q}{\varepsilon _o}\), \(\Sigma q\) will be +q1 only 

B Electric field at any point on this surface is only due to charge +q1

C Electric field at any point on this surface is due to charge +q1, – q2 and +q3

D \(\oint\limits \vec{E.}\,\,\vec{da}\) is the independent of the position of charge q1 inside the closed surface

×

For Gauss's law, flux is only dependent on the charge present inside the closed surface. 

Hence, \(\Sigma q\)  will be + q1 only. 

Hence, option (A) is correct.

Electric field at any point on the surface is not only due to charge present inside that surface, rather it is a property of space, i.e., electric field due to all charges present. 

Hence, option (B) is incorrect. 

 

Electric field at any point on the surface is due to all charges, because it is the property of space. 

Hence, option (C) is correct. 

For Gauss's law, flux is only dependent on the charge present inside the closed surface but does not depend on its position.

Hence, option (D) is correct. 

Consider a spherical Gaussian surface, as shown in figure. Choose the incorrect option.

image
A

For Gauss's law  \(\oint \vec{E.}\,\vec{da} = \dfrac {\Sigma q}{\varepsilon _o}\)\(\Sigma q\) will be +q1 only 

.

B

Electric field at any point on this surface is only due to charge +q1

C

Electric field at any point on this surface is due to charge +q1, – q2 and +q3

D

\(\oint\limits \vec{E.}\,\,\vec{da}\) is the independent of the position of charge q1 inside the closed surface

Option B is Correct

Electric Field at a Point Near Infinite Charged Body 

  • Consider an infinitely charged rod, as shown in figure. 

  • To Calculate electric field at a point P near this charged rod, take small elements of charge dq, as shown in figure.  
  • Electric field is produced at point P due to these two small charged elements having charge dq.
  • Net electric field at point P can be calculated by taking components. 
  • Vertical components of electric field cancel each other and horizontal components get added.
  • Hence, net electric field due to infinitely charged rod is in the direction perpendicular to the rod.

  • Any point near an infinite object is considered to be at the mid point of that object.So, the number of elements above the point is equal to the number of elements below the point.
  • Hence, net electric field at P will be perpendicular to the rod. 

  • Since, every point near infinite object is considered as a mid-point so, net electric field near infinite charged rod are represented, as shown in figure.

The electric field lines of infinite charged rod is shown in figure.

The top view of electric field lines of infinite charged rod is shown in figure. 

Similarly, for infinite charged sheet electric field are represented, as shown in figure.

The edge view of infinitely charged sheet is shown in figure.

Illustration Questions

Choose the correct option which represents the electric field lines of an infinite charged rod ( just near the rod ).

A

B

C

D

×

Net electric field due to infinitely charged rod is in the direction perpendicular to the rod.

Since, electric field lines are making an angle \(\theta\) with horizontal. Hence option (A) is incorrect.

image

Since, electric field is parallel to the rod. Hence, option (B) is incorrect .

image

Since, electric field lines are parallel to the rod. Hence, option (C) is incorrect.

image

Since, electric field is in the direction perpendicular to the rod. Hence, option (D) is correct.

image

Choose the correct option which represents the electric field lines of an infinite charged rod ( just near the rod ).

A image
B image
C image
D image

Option D is Correct

Gauss's Law 

  • Flux linked with any closed surface is given by Gauss's law as

\(\oint \vec{E.}\,\vec{dA} = \dfrac{\Sigma q}{\varepsilon_0}\)

where, \(\Sigma q\) represents summation of all charges enclosed by closed surface. 

  • By using Gauss's law, electric field can be calculated easily by choosing correct Gaussian surface.  

Steps to Choose the Correct Gaussian Surface

1. Selection of Gaussian surface should be such that the electric field and area vector are either parallel or perpendicular to each other. 

     Case 1: When area vector is perpendicular to field vector

     \(\oint \vec{E.}\;\vec{dA} = 0\)

     Case 2: When area vector is parallel to field vector 

     \(\oint \vec{E.}\;\vec{dA} = \oint E\, dA\)

2. The value of electric field should be constant over the whole surface.

     \(\oint \vec{E.}\;\vec{dA} = E\oint dA\)

3. Selection of surface should be such that the total surface area can be evaluated by 

     \(A = \oint d A\)

 

 

 

Illustration Questions

Choose the correct option of  selection of Gaussian surface to calculate electric field at a point outside a uniformly charged sphere.

A

B

C

D

×

For uniformly charged sphere, selection of Gaussian surface should be spherical because due to its symmetry, electric field is constant at all points on Gaussian surface.

Since, the Gaussian surface is cylindrical so, electric field is not constant over its whole surface. Hence,option (A) is incorrect. 

image

Since, the Gaussian surface is cylindrical and some charge of sphere is enclosed by it. Hence, option (B) is incorrect. 

image

Since, the Gaussian surface is selected such that it includes all the charges of sphere. Hence, option (C) is correct. 

image

Since, the Gaussian surface is cube and electric field is not constant over its whole surface. Hence, option (D) is incorrect. 

image

Choose the correct option of  selection of Gaussian surface to calculate electric field at a point outside a uniformly charged sphere.

A image
B image
C image
D image

Option C is Correct

Electric Field of a Charged Spherical Shell

  • Consider a spherical charged shell of radius R0, having charge +q.

Case 1:     Electric Field at a Point Inside Charged Spherical Shell 

  • To calculate electric field inside the sphere at point P situated at a distance r ( r < R0 ) from its center, consider a spherical Gaussian surface such that point P lies on it.
  • By application of Gauss's law for the Gaussian surface 

           \(\oint \vec{E.} \,\vec{dA}= \dfrac {\Sigma q}{\varepsilon _0}\)

  • Since, charge for the spherical shell resides on the surface hence, the charge enclosed by the Gaussian surface is zero.
  • So, net flux through Gaussian surface is zero.  

          \(\oint \vec{E.} \,\vec{dA} = 0\)

  • The Gaussian surface is chosen by symmetry in which the electric field vector is always parallel to the area vector. So, in this case the dot product of both the vector is non-zero.
  • Also, the area of the Gaussian surface is also non-zero. 
  • Hence, as flux through Gaussian surface is zero, it means electric field inside the spherical shell is zero.

            \(\vec{E} _{inside} = 0\)  

Case 2:      Electric Field at a Point Outside Charged Spherical Shell

  • To calculate electric field at a point S outside spherical shell at a distance R ( R \(\geq\) R0 ), consider a spherical Gaussian surface such that point S lies on it.  
  • By application of Gauss's law for Gaussian surface

           \(\oint \vec{E.}\;\vec{dA} = \dfrac {\Sigma q}{\varepsilon _0}\)

  • Since, total charge enclosed by Gaussian surface is q

         So,     \(\oint \vec{E.}\; \vec{dA} = \dfrac {q}{\varepsilon _0}\)

  • Also the symmetry of Gaussian surface is chosen such that electric field at all points on Gaussian surface is same and the direction of field is parallel to area vector.
  •         So, \(\oint \vec{E.}\;\vec{dA} = \dfrac {q}{\varepsilon_0}\)

                  \(\oint EdA = \dfrac {q}{\varepsilon _0}\)

            or,  \( E\oint dA = \dfrac {q}{\varepsilon _0}\)

            or,  \(E (4\;\pi R^{2}) = \dfrac{q}{\varepsilon_0}\)  { Area of spherical shell =\(4 \pi R^{2}\)}

            or,  \(E = \dfrac {1}{4 \pi R ^{2}} \dfrac{q}{\varepsilon _0}\)

                 \(E = \dfrac {1}{4 \pi \varepsilon_0} \dfrac{q}{R^{2}}\)    

  • Also the electric field at a distance R due to point charge is given as

           \(E = \dfrac {1}{4\pi \varepsilon_0} \dfrac{q}{R^{2}}\)    

Conclusion:

It can be concluded that electric field due to a spherical shell at an outside point is same as electric field due to point charge at distance R.

Illustration Questions

Calculate electric field due to a spherical shell having charge \(q = 9 \;nC\) and radius \(r = 3 \;cm\) at a point P at a distance \(R = 9 \; cm\) from its center.

A 104 V/m

B 10 V/m

C Zero

D None of these

×

Electric field due to spherical shell at outside point is given as \(\vec{E} = \dfrac{1}{4 \pi \varepsilon_0}\dfrac {q}{R^{2}}\)

Given : R = 9 cm  , q = 9 nC 

\(\vec{E} = \dfrac{9 × 10 ^{9}× 9 × 10 ^{-9}}{(9× 10 ^{-2})^2}\)

\(\vec{E} = 10^{4}\,\,V/m\)

Calculate electric field due to a spherical shell having charge \(q = 9 \;nC\) and radius \(r = 3 \;cm\) at a point P at a distance \(R = 9 \; cm\) from its center.

A

104 V/m

.

B

10 V/m

C

Zero

D

None of these

Option A is Correct

Gauss's Law and Selection of Gaussian Surface

Gauss's Law :

  • Flux linked with any closed surface is given by Gauss's law as \(\oint \vec{E.}\,\vec{dA} = \dfrac{\Sigma q}{\varepsilon_0}\) 

           where, \(\Sigma q\) represents summation of all charges enclosed by closed surface. 

  • By using Gauss's law, electric field can be calculated easily by choosing correct Gaussian surface.  

 Steps to Choose the Correct Gaussian Surface

  1. Selection of Gaussian surface should be such that the electric field and area vector are either parallel or perpendicular to each other. 

      Case 1. When area vector is perpendicular to field vector

      \(\oint \vec{E.}\;\vec{dA} = 0\)

      Case 2. When area vector is parallel to field vector 

      \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)

  2. The value of electric field should be constant over the whole surface.

     \(\oint \vec{E.}\;\vec{dA} = E\oint dA\)

  3. Selection of surface should be such that the total surface area can be evaluated by 

    \(A = \oint d A\)

 

Some Examples of Choice of Surface

(A) Point Charge 

  • For point charge, Gaussian surface should be spherical such that point charge is placed at the center of the spherical surface.
  • \(\vec{E}\) and \(\vec{dA}\) are parallel at every point.

So,   \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)

  • By symmetry, magnitude of electric field |\(\vec{E}\)| is same at all points on Gaussian spherical surface.

So,     \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)     

(B) Spherical Shell 

  • For spherical shell, Gaussian surface should be such that it includes all the charges inside it.
  • \(\vec{E}\) and \(\vec{dA}\) are parallel at every point.
  • So,   \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)

  • By symmetry, magnitude of electric field |\(\vec{E}\)| is same at all points on Gaussian spherical surface.
  • So,     \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)     

(C) Spherical Charge Distribution

  • \(\vec{E}\) and \(\vec{dA}\) are parallel at every point.

So,   \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)

  • By symmetry, magnitude of electric field |\(\vec{E}\)| is same at all points on Gaussian spherical surface.

So,     \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)   

(D) Long Charged Rod

  • Selection of Gaussian surface for a long charged rod should be cylindrical. 
  • \(\vec{E}\) and \(\vec{dA}\) are parallel at every point.
  • So,   \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)

  • By symmetry, magnitude of electric field |\(\vec{E}\)| is same at all points on Gaussian spherical surface.
  • So,     \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)   

The top view of electric field lines for long charged rod is shown in figure.

(E) Charged Sheet

  • Selection of Gaussian surface for a charged sheet is chosen to be cylindrical. 
  • For curved surface, electric field is perpendicular to the area vector.

     \(\oint \vec{E.}\;\vec{dA} = 0\)

  • For cross - section (end parts), electric field is parallel to the area vector.
  • \(\vec{E}\) and \(\vec{dA}\) are parallel at every point.

    So,   \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)

  • By symmetry, magnitude of electric field |\(\vec{E}\)| is same at all points on Gaussian spherical surface.
  • So,     \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)   

Illustration Questions

Choose the incorrect step for selection of Gaussian surface to calculate electric field due to point charge by Gauss's law.

A Choose spherical Gaussian surface such that point charge lies on its surface

B Choose spherical Gaussian surface such that point charge lies at its center

C \(\vec{E}\) and \(\vec{dA}\) are parallel for spherical Gaussian surface

D None of these

×

If point charge lies on spherical Gaussian surface then electric field will be different at each point on Gaussian surface and is non-uniform.

Hence, option (A) is incorrect.

If point charge lies at center of spherical Gaussian surface then electric field at each point on Gaussian surface is same and electric field vector is parallel to area vector.

Hence, option (B) is correct.

For spherical Gaussian surface, at every point on surface , direction of field vector is in same direction to area vector.

Hence, option (C) is correct. 

Choose the incorrect step for selection of Gaussian surface to calculate electric field due to point charge by Gauss's law.

A

Choose spherical Gaussian surface such that point charge lies on its surface

.

B

Choose spherical Gaussian surface such that point charge lies at its center

C

\(\vec{E}\) and \(\vec{dA}\) are parallel for spherical Gaussian surface

D

None of these

Option A is Correct

Application of Gauss's law for Concentric Charged Shells and Cylinders

  • Consider two concentric spherical shells of radius 'a' and 'b', as shown in figure. 
  • The inner sphere is given a charge +Q1 and the outer sphere is given a charge +Q2.
  • Calculation of electric field at any point due to this combination can be done by assuming a Gaussian surface passing through that point.  

Case 1: Electric field at point B ( r < a )

  • Selection of Gaussian surface should be such that point B lies on it.

  • The symmetry of Gaussian surface selected, is such that the electric field vector is in direction of area vector.

  • Since, electric field vector is in direction of area vector so, flux linked with Gaussian surface is given as

\(\phi = \oint \vec{E.}\;\vec{dA}\;\;\;\;\;[E\;||\;dA]\)

\(\phi = \oint {E}\;{dA}\)

\(\phi = E\oint \;{dA}\)       [Since| \(\vec{E}\) | is same at all points on Gaussian surface]

\(\phi =E(4 \pi r^2)\;\;\;\;\; —(1)\)        [Area of spherical Gaussian surface = \(4 \pi r ^2\)]

  • By using Gauss's law

\(\phi = \dfrac{\sum q}{\varepsilon_0} = 0\) [ Charge enclosed by Gaussian surface = 0 ]

So,    \(E (4 \pi r^{2}) = 0\)

means    E = 0

  • Hence, electric field at point B

\(\vec{E_B}_{net } = 0\)

Case 2. Electric Field at Point A at Distance r from Center ( a < r < b )

  • Selection of Gaussian surface is such that point A lies on it.

  • The symmetry of Gaussian surface selected, is such that the electric field vector is in direction of area vector.

  • Since, electric field vector is in direction of area vector. So, flux linked with Gaussian surface is given as.

\(\phi = \oint \vec {E.} \;\vec{dA}\)                             \( [\, E\, || \,dA \,]\)

\(\phi = \oint {E} \;{dA}\)

\(\phi = {E}\oint \;{dA}\)                        [ Since |\(\vec{E}\)| is same at all points on Gaussian surface ]

\(\phi = E (4 \pi r ^{2})\)  -------------(1)                    [ Area of spherical Gaussian surface = \(4 \pi r^{2}\)]  

  • By using Gauss's law

\(\phi = \dfrac{Q_1}{\varepsilon_0}\)  ---------(2)              [ Charge enclosed by Gaussian surface = Q1 ] 

From (1) and (2)

\(E \;\;(4 \pi r^{2}) = \dfrac{Q_1}{\varepsilon_0}\)

\(\vec{E} = \dfrac{1}{4\pi \varepsilon _0}\;\;\dfrac{Q_1}{r^{2}}\)    (Radially outwards)

Case 3. Electric Field at Point C ( r > b )

  • Selection of Gaussian surface should  be such that point C lies on it. 

The symmetry of Gaussian surface selected, is such that the electric field vector is in direction of area vector.

  • Since, electric field vector is in direction of area vector so, flux linked with Gaussian surface is given as

\(\phi = \oint \vec{E.}\;\vec{dA}\;\;\;\;\;[E||dA]\)

\(\phi = \oint {E}\;{dA}\)

\(\phi = E\oint \;{dA}\)       [Since| \(\vec{E}\) | is same at all points on Gaussian surface]

\(\phi =E(4 \pi r^2)\;\;\;\;\; —(1)\)        [Area of spherical Gaussian surface = \(4 \pi r ^2\)]

  • By using Gauss's law

\(\phi = \dfrac{Q_1+Q_2}{\varepsilon_ 0}\)     [ Charge enclosed by Gaussian surface = Q1 + Q2 ]

  • So, electric field at point C is given as

\(\vec{E} = \dfrac{1}{4 \pi \varepsilon_0}\dfrac{Q_1 + Q_2}{r^2}\)      (radially outwards)

Illustration Questions

Figure shows two concentric spherical shells of radii r1 and r2 having positive charge q1 and q2 respectively. Point A, B and C are at a distance a, b, and c respectively from center of the shell. Choose incorrect option regarding electric field at these points. 

A \(|\vec{E_A}| = \dfrac{1}{4 \pi \varepsilon _0} \;\;\dfrac{q_1}{a^{2}}\)

B \(|\vec{E_A}| = 0\)

C \(|\vec{E_B}| = \dfrac{1}{4 \pi \varepsilon _0} \;\;\dfrac{q_1}{b^{2}}\)

D \(|\vec{E_C}| = \dfrac{1}{4 \pi \varepsilon _0} \;\;\dfrac{q_1 + q_2}{c^{2}}\)

×

Electric field at point A

By Gauss's law

Flux  \(\phi =\dfrac{\Sigma q}{\varepsilon_0}\)

 

 

\(\phi = 0\)

[ Since, no charge enclosed by Gaussian surface \(\sum q = 0\) ]

Hence, electric field at A is Zero

 \(|\vec{E_A}| = 0\)      

image

Electric field at point B

By Gauss's law 

Flux \(\phi = \dfrac{\Sigma q}{\varepsilon_0}\) 

\(\phi = \dfrac {q_1}{\varepsilon_0}\)

  [ Change enclosed by Gaussian surface = q1 ]

So,    \(E_B (4 \pi b^{2}) = \dfrac{q_1}{\varepsilon _0}\)  

\(|\vec{E_B}| = \dfrac{q_1}{4\pi\varepsilon_0 b^{2}}\)

 

 

image

Electric field at point C

By Gauss's law 

Flux \(\phi = \dfrac{\Sigma q}{\varepsilon_0}\)

\(\phi = \dfrac{q_1 + q_2}{\varepsilon_0}\)  [ Charge enclosed by Gaussian surface = q1 + q2 ]

So,   \(E_c (4\pi c^{2}) = \dfrac{q_1 + q_2}{\varepsilon^{2}}\)

\(|\vec{E_c}| = \dfrac{1}{4 \pi \varepsilon _0} \;\;\dfrac{q_1 + q_2}{c^{2}}\)

 

 

image

Figure shows two concentric spherical shells of radii r1 and r2 having positive charge q1 and q2 respectively. Point A, B and C are at a distance a, b, and c respectively from center of the shell. Choose incorrect option regarding electric field at these points. 

image
A

\(|\vec{E_A}| = \dfrac{1}{4 \pi \varepsilon _0} \;\;\dfrac{q_1}{a^{2}}\)

.

B

\(|\vec{E_A}| = 0\)

C

\(|\vec{E_B}| = \dfrac{1}{4 \pi \varepsilon _0} \;\;\dfrac{q_1}{b^{2}}\)

D

\(|\vec{E_C}| = \dfrac{1}{4 \pi \varepsilon _0} \;\;\dfrac{q_1 + q_2}{c^{2}}\)

Option A is Correct

Electric Field at Different Points Due to Two Concentric Spheres 

  • Consider two concentric spheres with inner sphere is non - conducting of radius 'a' , having charge Q2 and outer sphere is conducting shell of radius 'b' having charge Q1.

Case 1:  Electric Field at Point P ( r < a )

  • Selection of Gaussian surface should be such that point P lies on it .

  • The symmetry of Gaussian surface selected, is such that the electric field vector is in direction of area vector.

  • Since, electric field vector is in direction of area vector. So, flux linked with Gaussian surface is given as

\(\phi = \oint \vec {E.} \;\vec{dA}\)                           \( [ \,E \,|| \,dA\, ]\)

\(\phi = \oint {E} \;{dA}\)

\(\phi = {E}\oint \;{dA}\)                        [ Since |\(\vec{E}\)| is same at all points on Gaussian surface ]

\(\phi = E (4 \pi r ^{2})\)  -------------(1)                     [ Area of spherical Gaussian surface = \(4 \pi r^{2}\)]  

By Gauss's law \(\phi = \dfrac {\Sigma q}{\varepsilon _0}\)  

[   Charge enclosed by closed surface  \(= \dfrac{Q_2}{\dfrac{4}{3}\pi a ^{3}} \; × \dfrac{4}{3} \pi r ^{3}\, =\, \dfrac{Q_2}{a^{3}}\;r^{3}\)   ]

 \(\phi = \dfrac{Q_2r^{3}}{\varepsilon _0a^{3}}\)

  •  Electric field at point P

 \(E_p(4 \pi r^{2})= \dfrac{Q_2r^{3}}{\varepsilon_0a^{3}}\)

 \(|\vec{E_P}| = \dfrac{Q_2\;r}{4\;\pi\varepsilon_0\;a^3}\)      

Case 2:  Electric Field at Point Q ( a < r < b )

  • Selection of Gaussian surface should be such that point Q lies on it. 

  • The symmetry of Gaussian surface selected, is such that the electric field vector is in direction of area vector.

  • Since, electric field vector is in direction of area vector. So, flux linked with Gaussian surface is given as

\(\phi = \oint \vec {E.} \;\vec{dA}\)                        \( [\, E\, || \,dA \,]\)

\(\phi = \oint {E} \;{dA}\)

\(\phi = {E}\oint \;{dA}\)                        [ Since |\(\vec{E}\)| is same at all points on Gaussian surface ]

\(\phi = E (4 \pi r ^{2})\)  -------------(1)                     [ Area of spherical Gaussian surface = \(4 \pi r^{2}\)]  

By Gauss's law 

\(\phi = \dfrac {\Sigma q }{\varepsilon _0}\)

 [ Charge enclosed by Gaussian Surface = Q]

or,  \(\phi = \dfrac{Q_2}{\varepsilon_0}\)     

  • Electric Field at Point Q

\(E_Q(4\pi r^{2}) = \dfrac{Q_2}{\varepsilon_0}\)

\(|\vec{E_Q}| = \dfrac{Q_2}{4 \pi \varepsilon_0r^{2}}\)   

Case 3: Electric Field at Point R ( r > b )

  • Selection of Gaussian surface should be such that point R lies on it.

  • The symmetry of Gaussian surface selected, is such that the electric field vector is in direction of area vector.

  • Since, electric field vector is in direction of area vector. So, flux linked with Gaussian surface is given as

\(\phi = \oint \vec {E.} \;\vec{dA}\)                     \( [\, E\, || \,dA \,]\)

\(\phi = \oint {E} \;{dA}\)

\(\phi = {E}\oint \;{dA}\)                        [ Since |\(\vec{E}\)| is same at all points on Gaussian surface ]

\(\phi = E (4 \pi r ^{2})\)  -------------(1)                     [ Area of spherical Gaussian surface = \(4 \pi r^{2}\)]  

  • By using Gauss's law 

\(\phi = \dfrac{\Sigma q}{\varepsilon_0}\)         

[ Charge enclosed by Gaussian Surface = Q1 + Q2 ]

\(\phi = \dfrac{Q_1 + Q_2}{\varepsilon_0 }\)

  •  Electric field at point R

\(E_R (4\pi r^{2}) = \dfrac{Q_1 + Q_2}{\varepsilon_0}\)

\(|\vec{E_R}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{Q_1 +Q_2}{r^{2}}\)     

Illustration Questions

Figure shows two concentric spheres with inner sphere is non - conducting of radius r1 having charge +q1 and outer sphere is conducting of radius r2 having charge q2. Points A,B and C are at a distance a, b and c respectively, from center of shell. Choose incorrect option regarding electric field at these points.   

A \(|\vec{E_A}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_1a}{r_1^{3}}\)

B \(|\vec{E_B}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_1}{b^{2}}\)

C \(|\vec{E_C}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_2}{c^{2}}\)

D \(|\vec{E_C}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_1 +q_2}{c^{2}}\)

×

Electric Field at Point A 

By Gauss's law          

Flux  \(\phi = \dfrac {\Sigma q}{\varepsilon_0}\)   [ Charge enclosed by Gaussian surface   \(= \dfrac {q_1}{\dfrac{4}{3} \pi r_1^{3}} × 4/3\pi a^{3}\;\;= \dfrac{q_1a^{3}}{r_1^{3}}\)  ]

\(\phi= \dfrac {q_1 a^{3}}{\varepsilon _0r_1\;^{3}}\)    

Hence, electric field at A

\(E_A(4\pi a^{2}) = \dfrac {q_1a^{3}}{\varepsilon_0 r_1^{3}}\)

\(|\vec {E}_A| = \dfrac {1}{4\pi \varepsilon_0} \dfrac {q_1a}{r_1^{3}}\)

 

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Electric Field at Point B

By Gauss's law 

Flux  \(\phi = \dfrac{\Sigma q}{\varepsilon _0}\)             [ Charge enclosed by Gaussian Surface = q1 ]

\(\phi = \dfrac{q_1}{\varepsilon_0}\)

Electric field at B 

\(E_B (4\pi b ^{2}) = \dfrac{q_1}{\varepsilon _0{}{}}\)

\(|\vec E_B| = \dfrac{1}{4 \pi \varepsilon_0 } \dfrac{q_1}{b^{2}}\)

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Electric Field at Point C

By Gauss's law 

Flux  \(\phi = \dfrac{\Sigma q}{\varepsilon _0}\)             [ Charge enclosed by Gaussian Surface = q1 + q2 ]

\(\phi = \dfrac{q_1 + q_2}{\varepsilon_0}\)

Electric Field at C

\(E_c (4\pi c ^{2}) = \dfrac{q_1 + q_2}{\varepsilon _0{}{}}\)

\(|\vec E_c| = \dfrac{1}{4 \pi \varepsilon_0} \;\;\dfrac {q_1 +q_2}{c_2}\)

 

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Figure shows two concentric spheres with inner sphere is non - conducting of radius r1 having charge +q1 and outer sphere is conducting of radius r2 having charge q2. Points A,B and C are at a distance a, b and c respectively, from center of shell. Choose incorrect option regarding electric field at these points.   

image
A

\(|\vec{E_A}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_1a}{r_1^{3}}\)

.

B

\(|\vec{E_B}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_1}{b^{2}}\)

C

\(|\vec{E_C}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_2}{c^{2}}\)

D

\(|\vec{E_C}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_1 +q_2}{c^{2}}\)

Option C is Correct

Electric Field at Different Points due to Uniformly Charged Hollow Cylinder 

  • Consider a uniformly charged hollow cylinder of infinite length with a radius R, length L,( R << L ) having charge Q. 
  • Consider point P at a distance r1 and point P2 at a perpendicular distance r2 from the axis, as shown in figure.

Case 1: Electric Field at Point P1 ( r1 < R )

  • Selection of Gaussian  surface should be such that point P1, lies on its surface. 
  • Gaussian surface will be cylindrical of length \(l\) and radius r1 \([ l<<<L] \).
  • Electric Field at Point P1
  • Using Gauss's law 

    \(\phi = \oint \vec {E.}\ dA = \dfrac {q_{inside} }{\varepsilon _0}\)           [ Charge inside Gaussian surface = 0 ]

    \(\phi = 0 = \oint \vec{E.} \;\vec{dA}\)

    \( \oint \vec {E_{p_1}.}\;\vec{dA} = 0\)

    \( \oint {E_{p_1}.}\;{dA} = 0\)

    \( {E_{p_1}.} \oint\;{dA} = 0\)

    \( {E_{p_1}.} 2 \pi r_1l = 0\)

    \( {E_{p_1}} = 0\)    

Case 2: Electric Field at Point P2 ( r2 > R )

  • Selection of Gaussian surface should be such that point P2 lies on its surface.
  • Gaussian surface will be cylindrical of length  \( l\) and radius r2 \([ l<<<L]\).
  • Electric Field at Point P2 
  • Using Gauss's law 

    \(\phi = \oint \vec {E.}\ dA = \dfrac {q_{inside} }{\varepsilon _0}\)           [ Charge inside Gaussian surface  \(=\dfrac{Q}{L}× l\) ]

    \(\phi = \oint \vec{E.} \;\vec{dA}\)   =  \(\dfrac{Q l }{\varepsilon_0L}\)

    \(E\oint d A = \dfrac {Q l }{L\varepsilon_0}\)

    \( {E.}\; 2 \pi r_2l = \dfrac {Ql}{L\varepsilon_0}\)

    \(E = \dfrac {Q}{2 \pi r_2 L\varepsilon_0}\)

     

     

Illustration Questions

Figure shows a uniformly charged hollow cylinder of infinite length with radius R and charge Q. Points P1 and P2 are respectively r1 and r2 distance apart from the axis. Choose the incorrect option regarding electric field at these points. 

A \(E_{p_1} = \dfrac{Q}{2 \pi r_1L\varepsilon _0}\)

B \(E_{p_2} = \dfrac{Q}{2 \pi r_2L\varepsilon _0}\)

C \(E_{p_1} = 0\)

D

×

Electric Field at Point P1

By Gauss's law 

Flux  \(\phi = \dfrac {\Sigma q}{\varepsilon_0}\)

[ Charge enclosed by Gaussian surface = 0 ]   

\(\phi = 0\)

Hence, electric field

 \(\oint E.dA \;= E_{p_1} (2 \pi r_1l)= 0\)

\( E_{p_1} = 0\)

 

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Electric Field at Point P2

By Gauss's law 

Flux  \(\phi = \dfrac {\Sigma q}{\varepsilon_0}\)

[ Charge enclosed by Gaussian surface = \(\dfrac{Q}{L} × l\)  ]   

\(\phi = \dfrac{Q l }{L\varepsilon_0}\)

Electric Field at P2

\(E( 2 \pi r_2l) =\dfrac{Ql}{L\varepsilon_0}\)

\(E = \dfrac{Q}{2 \pi \varepsilon_0r_2L}\)   

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Figure shows a uniformly charged hollow cylinder of infinite length with radius R and charge Q. Points P1 and P2 are respectively r1 and r2 distance apart from the axis. Choose the incorrect option regarding electric field at these points. 

image
A

\(E_{p_1} = \dfrac{Q}{2 \pi r_1L\varepsilon _0}\)

.

B

\(E_{p_2} = \dfrac{Q}{2 \pi r_2L\varepsilon _0}\)

C

\(E_{p_1} = 0\)

D

Option A is Correct

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