Advertisements
Advertisements
प्रश्न
State Gauss’s law on electrostatics and drive expression for the electric field due to a long straight thin uniformly charged wire (linear charge density λ) at a point lying at a distance r from the wire.
Advertisements
उत्तर
Gauss' Law states that the net electric flux through any closed surface is equal to `1/epsilon_0` times the net electric charge within that closed surface.
`oint vec" E".d vec" s" = (q_(enclosed))/epsilon_o`
In the diagram, we have taken a cylindrical gaussian surface of radius = r and length = l.
The net charge enclosed inside the gaussian surface `q_(enclosed) = lambdal`
By symmetry, we can say that the Electric field will be in radially outward direction.
According to gauss' law,
`oint vec"E".d vec"s" = q_(enclosed)/epsilon_o`
`int_1 vec"E" .d vec"s" + int_2 vec"E" .d vec"s" + int_3 vec"E". d vec"s" = (lambdal)/epsilon_o`
`int_1 vec"E". d vec"s" & int_3 vec"E". d vec"s" "are zero", "Since" vec"E" "is perpendicular to" d vec"s"`
`int_2 vec"E" . d vec"s" = (lambdal)/epsilon_o`
`"at" 2, vec"E" and d vec"s" "are in the same direction, we can write"`
`E.2pirl = (lambdal)/epsilon_o`
`E = lambda/(2piepsilon_o r)`
APPEARS IN
संबंधित प्रश्न
State and explain Gauss’s law.
A charge Q is placed at the centre of a cube. Find the flux of the electric field through the six surfaces of the cube.
q1, q2, q3 and q4 are point charges located at points as shown in the figure and S is a spherical gaussian surface of radius R. Which of the following is true according to the Gauss' law?
The Gaussian surface ______.
The surface considered for Gauss’s law is called ______.
Which of the following statements is not true about Gauss’s law?
The Electric flux through the surface
 (i) |
 (ii) |
 (iii) |
 (iv) |
Refer to the arrangement of charges in figure and a Gaussian surface of radius R with Q at the centre. Then
- total flux through the surface of the sphere is `(-Q)/ε_0`.
- field on the surface of the sphere is `(-Q)/(4 piε_0 R^2)`.
- flux through the surface of sphere due to 5Q is zero.
- field on the surface of sphere due to –2Q is same everywhere.
The region between two concentric spheres of radii a < b contain volume charge density ρ(r) = `"c"/"r"`, where c is constant and r is radial- distanct from centre no figure needed. A point charge q is placed at the origin, r = 0. Value of c is in such a way for which the electric field in the region between the spheres is constant (i.e. independent of r). Find the value of c:
