Advertisements
Advertisements
प्रश्न
State Gauss’ Law.
Answer the following question briefly and to the point:
State Gauss’ theorem.
Advertisements
उत्तर १
The electric flux (ΦE) through any closed surface is equal to `1/in_0` times the ‘net’ change q enclosed by the surface.
ΦE = `oint vec E d vec A`
= `q/in_0`
∈0 = Permittivity of free space.
उत्तर २
Gauss’ theorem states that the net electric flux over a closed surface is `1/epsilon_0` times the net electric charge enclosed by the surface.
Φ = `oint vec E * d vec A`
= `q/epsilon_0`
संबंधित प्रश्न
Explain why, for a charge configuration, the equipotential surface through a point is normal to the electric field at that point
A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
A point charge causes an electric flux of −1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
- If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
- What is the value of the point charge?
A point charge q is at a distance of d/2 directly above the centre of a square of side d, as shown the figure. Use Gauss' law to obtain the expression for the electric flux through the square.
Use Gauss' law to derive the expression for the electric field `(vecE)` due to a straight uniformly charged infinite line of charge density λ C/m.
Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
The electric field intensity outside the charged conducting sphere of radius ‘R’, placed in a medium of permittivity ∈ at a distance ‘r’ from the centre of the sphere in terms of surface charge density σ is
Two wires A and B of the same material and of equal length with the radii in the ratio 1 : 2 are subjected to identical loads. If the length of A increases by 8 mm, then the increase in length of B is:
Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure.
