The electric field outside the metal sphere is
Answers
Answered by
1
Hey.
Here is the answer.
The electric field outside the metal sphere is diverging lines or rays emerging from the center of sphere as if whole charge is concentrated at the center .
Explanation :
Let us take a spherical Gaussian surface of radius R (where,R>r). r is radius of hollow sphere. If E is field at R, then according to Gauss’s theorem surface integral E.da over Gaussian surface is equal to charge Q enclosed by the Gaussian surface divided by epsilon zero.
Here,da is surface element or area vector.
But, E is constant on our chosen Gaussian surface. Therefore E can be taken out from surface integral and then integral da over the surface is
4 π R^2.
Thus, E 4 π R^2 = Q / (epsilon zero)
Or
E = Q/ 4 π epsilon zero R^2.
This is the field out side the given hollow sphere. This field is as if the whole charge on the outer surface is concentrated on the center of the hollow sphere.
Thanks.
Here is the answer.
The electric field outside the metal sphere is diverging lines or rays emerging from the center of sphere as if whole charge is concentrated at the center .
Explanation :
Let us take a spherical Gaussian surface of radius R (where,R>r). r is radius of hollow sphere. If E is field at R, then according to Gauss’s theorem surface integral E.da over Gaussian surface is equal to charge Q enclosed by the Gaussian surface divided by epsilon zero.
Here,da is surface element or area vector.
But, E is constant on our chosen Gaussian surface. Therefore E can be taken out from surface integral and then integral da over the surface is
4 π R^2.
Thus, E 4 π R^2 = Q / (epsilon zero)
Or
E = Q/ 4 π epsilon zero R^2.
This is the field out side the given hollow sphere. This field is as if the whole charge on the outer surface is concentrated on the center of the hollow sphere.
Thanks.
Similar questions