Gauss's law in electrostatics explains the relation between flux related to a surface and total charge enclosed by the surface. To derive the relation, consider a charge 'q' kept at the centre of a spherical surface of radius 'r'.
a) The magnitude of electric field on the surface.
b) We can consider that the spherical surface is made up of a number of small plane area each equal to 'dS'. The angle between the direction of electric field vector and area vector is zero.
c) The electric flux through such an area element ds
d) The electric flux through the entire spherical surface.
ie.. Total flux over a closed surface in free space is(one/epsilone) times the total charge enclosed by the surface. This is called Gauss's theorem.
If there are several charges q1,q2,q3,. inside the closed surface, each will contribute to the total electric flux. Total electric flux is the summation * the 1/epsilon
Where epsilon q is the algebraic sum of the charges within the surfaces.
Note:
1) Since the electric lines are emerging from the charge 'q' the electric flux will not change even if the spherical surface of arbitrary shape. So an appropriate Gaussian surface is selected based on the symmetry of the problem under consideration. The examples illustrated in the coming section demonstrates the choice of Gaussian surface.
2) Now, as far as the charge is anywhere within the closed surface, the electric flux will not change. So we can say that the total flux through a closed surface is 1/epsilonzero times charge enclosed by the surface.
3) But if we keep an electric dipole within the closed surface of arbitrary shape, net flux from the closed surface is zero. The is because the electric lines emerging from positive charge ends on negative charge. Hence no net outward or inward flux, i.e, when the net charge inside the surface is zero, the net flux through a closed surface is zero. If the charge enclosed is positive, the flux is outward and if it is negative, flux is inward.
a) The magnitude of electric field on the surface.
b) We can consider that the spherical surface is made up of a number of small plane area each equal to 'dS'. The angle between the direction of electric field vector and area vector is zero.
c) The electric flux through such an area element ds
d) The electric flux through the entire spherical surface.
ie.. Total flux over a closed surface in free space is(one/epsilone) times the total charge enclosed by the surface. This is called Gauss's theorem.
If there are several charges q1,q2,q3,. inside the closed surface, each will contribute to the total electric flux. Total electric flux is the summation * the 1/epsilon
Where epsilon q is the algebraic sum of the charges within the surfaces.
Note:
1) Since the electric lines are emerging from the charge 'q' the electric flux will not change even if the spherical surface of arbitrary shape. So an appropriate Gaussian surface is selected based on the symmetry of the problem under consideration. The examples illustrated in the coming section demonstrates the choice of Gaussian surface.
2) Now, as far as the charge is anywhere within the closed surface, the electric flux will not change. So we can say that the total flux through a closed surface is 1/epsilonzero times charge enclosed by the surface.
3) But if we keep an electric dipole within the closed surface of arbitrary shape, net flux from the closed surface is zero. The is because the electric lines emerging from positive charge ends on negative charge. Hence no net outward or inward flux, i.e, when the net charge inside the surface is zero, the net flux through a closed surface is zero. If the charge enclosed is positive, the flux is outward and if it is negative, flux is inward.
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