Advertisements
Advertisements
Question
A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s law, derive an expression for an electric field at a point outside the shell.
Advertisements
Solution
According to Gauss law,
`epsi_0EointdA=q`
Where,
q is the point charge
E is electric field due to the point charge
dA is a small area on the Gaussian surface at any distance and
`epsi_0` is the proportionality constant
For a spherical shell at distance r from the point charge, the integral `ointdA` is merely the sum of all differential of dA on the sphere.
Therefore, `oint dA =4pir^2`
`epsi_0E(4pir^2) = q`
or `,E = q/(epsi4pir^2)`
Therefore, for a thin conducting spherical shell of radius R and charge Q, spread uniformly over its surface, the electric field at any point outside the shell is
`E = Q/(e_0 4pir^2)`
Where r is the distance of the point from the centre of the shell.
`E = q/(4piepsi_0r^2)`
APPEARS IN
RELATED QUESTIONS
A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in the Figure. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)
A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube?
Draw a graph of electric field E(r) with distance r from the centre of the shell for 0 ≤ r ≤ ∞.
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.
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?
Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region
- the electric field is necessarily zero.
- the electric field is due to the dipole moment of the charge distribution only.
- the dominant electric field is `∞ 1/r^3`, for large r, where r is the distance from a origin in this region.
- the work done to move a charged particle along a closed path, away from the region, will be zero.
In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep = – (1 + y)e where e is the electronic charge.
- Find the critical value of y such that expansion may start.
- Show that the velocity of expansion is proportional to the distance from the centre.
In finding the electric field using Gauss law the formula `|vec"E"| = "q"_"enc"/(epsilon_0|"A"|)` is applicable. In the formula ε0 is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?
A charge Q is placed at the centre of a cube. The electric flux through one of its faces is ______.
In the diagram, the total electric flux through the closed surface ‘S’ is:
[Given q = charge ε0 = permittivity of free space]
