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1800-102-2727It is quite interesting that inside of a spherical shell the electric field intensity is zero. If you measure the potential inside a spherical shell there will be some non-zero value and it will be constant throughout. Actually all the charge a shell contains comes to the surface and thus no charge is left inside. Thus Gauss Law is really helpful in calculating the field intensity at any point. Knowing the charge distribution is a key to application of Gauss law.
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Gauss law can be used in determining the electric field intensity inside or outside a uniformly charged spherical shell. In case we are to determine the electric field outside the charged spherical shell, we consider an imaginary sphere passing through the point having r>R. And in case we are to determine the electric field inside the charged spherical shell, we consider an imaginary sphere passing through the point having r<R.
Calculation of electric field inside a charged spherical shell:
A spherical shell of radius R is charged positively so that its surface charge density is
We are to calculate the electric field intensity at a radial distance r inside the sphere i.e., r<R.
We imagine a spherical surface of radius r. Let the electrical field intensity on the imaginary spherical surface be E.
The charge enclosed by that imaginary spherical surface is 0 as all the charge resides on the sphere.
According to the Gauss law,
$i.e.,\frac{0}{{\epsilon}_{0}}=E\times 4\pi {r}^{2}\Rightarrow E=0$
So the electric field inside a charged spherical shell is zero.
Calculation of electric field outside a charged spherical shell:
A spherical shell of radius R is charged positively so that its surface charge density is .
We are to calculate the electric field intensity at a radial distance r outside the sphere i.e., r>R.
We imagine a spherical surface of radius r. Let the electrical field intensity on the imaginary spherical surface be E.
The charge enclosed by that imaginary spherical surface is σ×4πR2 as all the charge resides on the sphere.
According to the Gauss law,
$i.e.,\frac{\sigma \times 4\pi {R}^{2}}{{\epsilon}_{0}}=E\times 4\pi {r}^{2}\Rightarrow E=\frac{\sigma {R}^{2}}{{\epsilon}_{0}{r}^{2}}$
$E=\frac{\sigma {R}^{2}}{{\epsilon}_{0}{r}^{2}}i.e.,E\propto \frac{1}{{r}^{2}}outsidethesphere$
Q. A spherical shell of radius 1 m with surface charge density 8 Cm-2 is at rest with the centre at the origin. The Electric field at (0.5 m,0.5 m, 0.5 m) is?
A. As the sphere is of radius 1 m the point (0.5 m,0.5 m, 0.5 m) refers to the inside of the spherical shell. This point is on the surface of an imaginary sphere of radius $\sqrt{{0.5}^{2}+{0.5}^{2}+{0.5}^{2}}=0.87m$
The whole charge is on the surface of the sphere. So inside it no charge resides.
According to the Gauss law,
As,${q}_{enclosed}=0\Rightarrow E=0$
Q. A spherical shell of radius 1 m with surface charge density 0.5 Cm-2 is at rest with the centre at the origin. The Electric field at (2 m,2 m,2 m) is?
A. As the sphere is of radius 1 m the point (2 m,2 m,2 m) refers to the outside of the spherical shell. This point is on the surface of an imaginary sphere of radius $\sqrt{{2}^{2}+{2}^{2}+{2}^{2}}=3.4641m$
The whole charge is on the surface of the sphere. So inside the imaginary sphere of radius 3.464 m it has charge 4π12×0.5 =2π C.
According to the Gauss law,
qenclosed0=E.dA
$\Rightarrow \frac{2\pi}{{\epsilon}_{0}}=E\times 4\pi \left(3\times {2}^{2}\right)\Rightarrow E=\frac{1}{24{\epsilon}_{0}}N{C}^{-1}$
Q. A spherical shell of radius 1.5 m having charge of 4 C is at rest with the centre at the origin. The Electric field at (23 m,23 m,23 m) is
A. This point is on the surface of an imaginary sphere of radius $\sqrt{{\left(\frac{2}{\sqrt{3}}\right)}^{2}+{\left(\frac{2}{\sqrt{3}}\right)}^{2}+{\left(\frac{2}{\sqrt{3}}\right)}^{2}}=2m$
As the sphere is of radius 1.5 m the point $(\frac{2}{\sqrt{3}}m,\frac{2}{\sqrt{3}}m,\frac{2}{\sqrt{3}}m)$ refers to the outside of the spherical shell.
The whole charge is on the surface of the sphere. So inside the imaginary sphere of radius 2 m it has charge 4C.
According to the Gauss law,
$\Rightarrow \frac{4}{{\epsilon}_{0}}=E\times 4\pi {2}^{2}\Rightarrow E=\frac{1}{4\pi {\epsilon}_{0}}=9\times {10}^{9}N{C}^{-1}$
Q. A spherical shell of radius 1 m with surface charge density 35.41610-12 Cm-2 is at rest with the centre at the origin. The Electric field at a radial distance of 2 m is?
A. As the sphere is of radius 1 m, the point at a radial distance of 2 m refers to the outside of the spherical shell. This point is on the surface of an imaginary sphere of radius 2 m.
The whole charge is on the surface of the sphere. So inside the imaginary sphere of radius 2 m it has charge 4π12×35.41610-12 C.
According to the Gauss law,
$\Rightarrow \frac{4\pi {1}^{2}\times 35.416\times {10}^{-12}}{{\epsilon}_{0}}=E\times 4\pi {2}^{2}\Rightarrow E=1N{C}^{-1}$
Q. Explain without the application of Gauss law, how the electric field intensity at the centre of a metallic spherical shell is zero.
A. Let us consider a spherical shell where it has a charge. The charge will spread over the surface instantly. Now because of a spherical geometry the charge distribution will be even throughout the surface.
If we consider the centre, it will have an electrical field which is due to a vector summation of all the distributed point charges on the surface.
Because of symmetry this electrical intensity at the centre will be the same in magnitude but radially in different directions. The electric field intensity due to an infinitesimally small charged area will be canceled by the similar one positioned diametrically opposite to it. Thus the vector sum will be zero.
Q. Which of the graphs accurately shows the variation of the electric field for a charged spherical shell?
A. a
The field inside is zero and it decreases with the second power of the distance from the centre.
Q. Does charge distribution depend on whether the conducting sphere is hollow or solid?
A. No, since the charges are repelled and reside on the surface, the charge distribution is effectively the same in both cases.
Q. Can Gauss law be used if the charge distribution on the shell is non uniform?
A. In this case the electric field lines will not be symmetrical, hence using Gauss law will not help much.