Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
State and explain Gauss’s law.
Answer the following question.
State Gauss's law for magnetism. Explain its significance.
Gaussian surface cannot pass through discrete charge because ____________.
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 ______.
Five charges q1, q2, q3, q4, and q5 are fixed at their positions as shown in figure. S is a Gaussian surface. The Gauss’s law is given by `oint_s E.ds = q/ε_0`
Which of the following statements is correct?
If `oint_s` E.dS = 0 over a surface, then ______.
- the electric field inside the surface and on it is zero.
- the electric field inside the surface is necessarily uniform.
- the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
- all charges must necessarily be outside the surface.
A charge of +5 μC is placed at the centre of two concentric spheres of radii r1 = 3 cm and r2 = 5 cm. The ratio of the flux through sphere of radius r1 to that through sphere of radius r2 will be ______.
