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.
\(\phi_E = \int\limits _S \vec{E.}\,\,\vec{da}\)
\(\phi \propto q_{in}\)
or, \(\phi = \dfrac{q_{in}}{\varepsilon _0}\)
where \(\varepsilon _0\) is the permittivity of the free space.
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
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.
\(\oint \vec{E.}\,\vec{dA} = \dfrac{\Sigma q}{\varepsilon_0}\)
where, \(\Sigma q\) represents summation of all charges enclosed by closed 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
\(A = \oint d A\)
Case 1: Electric Field at a Point Inside Charged Spherical Shell
\(\oint \vec{E.} \,\vec{dA}= \dfrac {\Sigma q}{\varepsilon _0}\)
\(\oint \vec{E.} \,\vec{dA} = 0\)
\(\vec{E} _{inside} = 0\)
Case 2: Electric Field at a Point Outside Charged Spherical Shell
\(\oint \vec{E.}\;\vec{dA} = \dfrac {\Sigma q}{\varepsilon _0}\)
So, \(\oint \vec{E.}\; \vec{dA} = \dfrac {q}{\varepsilon _0}\)
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, \(E = \dfrac {1}{4 \pi R ^{2}} \dfrac{q}{\varepsilon _0}\)
\(E = \dfrac {1}{4 \pi \varepsilon_0} \dfrac{q}{R^{2}}\)
\(E = \dfrac {1}{4\pi \varepsilon_0} \dfrac{q}{R^{2}}\)
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.
A 104 V/m
B 10 V/m
C Zero
D None of these
Gauss's Law :
where, \(\Sigma q\) represents summation of all charges enclosed by closed 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
\(A = \oint d A\)
(A) Point Charge
So, \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)
So, \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)
(B) Spherical Shell
So, \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)
So, \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)
(C) Spherical Charge Distribution
So, \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)
So, \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)
(D) Long Charged Rod
So, \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)
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
\(\oint \vec{E.}\;\vec{dA} = 0\)
\(\vec{E}\) and \(\vec{dA}\) are parallel at every point.
So, \(\oint \vec{E.}\;\vec{dA} = \oint E dA\)
So, \(\oint \vec{E.}\;\vec{dA} = E \oint dA\)
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
Case 1: Electric field at point B ( r < a )
\(\phi = \oint \vec{E.}\;\vec{dA}\;\;\;\;\;[E\;||\;dA]\)
\(\phi = \oint {E}\;{dA}\)
\(\phi = E\oint \;{dA}\) [Since| \(\vec{E}\) | is same at all points on Gaussian surface]
\(\phi =E(4 \pi r^2)\;\;\;\;\; —(1)\) [Area of spherical Gaussian surface = \(4 \pi r ^2\)]
\(\phi = \dfrac{\sum q}{\varepsilon_0} = 0\) [ Charge enclosed by Gaussian surface = 0 ]
So, \(E (4 \pi r^{2}) = 0\)
means E = 0
\(\vec{E_B}_{net } = 0\)
Case 2. Electric Field at Point A at Distance r from Center ( a < r < b )
\(\phi = \oint \vec {E.} \;\vec{dA}\) \( [\, E\, || \,dA \,]\)
\(\phi = \oint {E} \;{dA}\)
\(\phi = {E}\oint \;{dA}\) [ Since |\(\vec{E}\)| is same at all points on Gaussian surface ]
\(\phi = E (4 \pi r ^{2})\) -------------(1) [ Area of spherical Gaussian surface = \(4 \pi r^{2}\)]
\(\phi = \dfrac{Q_1}{\varepsilon_0}\) ---------(2) [ Charge enclosed by Gaussian surface = Q_{1} ]
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 )
The symmetry of Gaussian surface selected, is such that the electric field vector is in direction of area vector.
\(\phi = \oint \vec{E.}\;\vec{dA}\;\;\;\;\;[E||dA]\)
\(\phi = \oint {E}\;{dA}\)
\(\phi = E\oint \;{dA}\) [Since| \(\vec{E}\) | is same at all points on Gaussian surface]
\(\phi =E(4 \pi r^2)\;\;\;\;\; —(1)\) [Area of spherical Gaussian surface = \(4 \pi r ^2\)]
\(\phi = \dfrac{Q_1+Q_2}{\varepsilon_ 0}\) [ Charge enclosed by Gaussian surface = Q_{1} + Q_{2} ]
\(\vec{E} = \dfrac{1}{4 \pi \varepsilon_0}\dfrac{Q_1 + Q_2}{r^2}\) (radially outwards)
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 \(|\vec{E_C}| = \dfrac{1}{4 \pi \varepsilon _0} \;\;\dfrac{q_1 + q_2}{c^{2}}\)
Case 1: Electric Field at Point P ( r < a )
\(\phi = \oint \vec {E.} \;\vec{dA}\) \( [ \,E \,|| \,dA\, ]\)
\(\phi = \oint {E} \;{dA}\)
\(\phi = {E}\oint \;{dA}\) [ 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 \(= \dfrac{Q_2}{\dfrac{4}{3}\pi a ^{3}} \; × \dfrac{4}{3} \pi r ^{3}\, =\, \dfrac{Q_2}{a^{3}}\;r^{3}\) ]
\(\phi = \dfrac{Q_2r^{3}}{\varepsilon _0a^{3}}\)
\(E_p(4 \pi r^{2})= \dfrac{Q_2r^{3}}{\varepsilon_0a^{3}}\)
\(|\vec{E_P}| = \dfrac{Q_2\;r}{4\;\pi\varepsilon_0\;a^3}\)
Case 2: Electric Field at Point Q ( a < r < b )
\(\phi = \oint \vec {E.} \;\vec{dA}\) \( [\, E\, || \,dA \,]\)
\(\phi = \oint {E} \;{dA}\)
\(\phi = {E}\oint \;{dA}\) [ 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_{2 }]
or, \(\phi = \dfrac{Q_2}{\varepsilon_0}\)
\(E_Q(4\pi r^{2}) = \dfrac{Q_2}{\varepsilon_0}\)
\(|\vec{E_Q}| = \dfrac{Q_2}{4 \pi \varepsilon_0r^{2}}\)
Case 3: Electric Field at Point R ( r > b )
\(\phi = \oint \vec {E.} \;\vec{dA}\) \( [\, E\, || \,dA \,]\)
\(\phi = \oint {E} \;{dA}\)
\(\phi = {E}\oint \;{dA}\) [ Since |\(\vec{E}\)| is same at all points on Gaussian surface ]
\(\phi = E (4 \pi r ^{2})\) -------------(1) [ Area of spherical Gaussian surface = \(4 \pi r^{2}\)]
\(\phi = \dfrac{\Sigma q}{\varepsilon_0}\)
[ Charge enclosed by Gaussian Surface = Q_{1} + Q_{2} ]
\(\phi = \dfrac{Q_1 + Q_2}{\varepsilon_0 }\)
\(E_R (4\pi r^{2}) = \dfrac{Q_1 + Q_2}{\varepsilon_0}\)
\(|\vec{E_R}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{Q_1 +Q_2}{r^{2}}\)
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 \(|\vec{E_C}| = \dfrac{1}{4\pi \varepsilon_0} \; \dfrac{q_1 +q_2}{c^{2}}\)
Case 1: Electric Field at Point P_{1} ( r_{1} < R )
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 P_{2} ( r_{2} > R )
Using Gauss's law
\(\phi = \oint \vec {E.}\ dA = \dfrac {q_{inside} }{\varepsilon _0}\) [ Charge inside Gaussian surface \(=\dfrac{Q}{L}× l\) ]
\(\phi = \oint \vec{E.} \;\vec{dA}\) = \(\dfrac{Q l }{\varepsilon_0L}\)
\(E\oint d A = \dfrac {Q l }{L\varepsilon_0}\)
\( {E.}\; 2 \pi r_2l = \dfrac {Ql}{L\varepsilon_0}\)
\(E = \dfrac {Q}{2 \pi r_2 L\varepsilon_0}\)
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