Question

What is the electric field due to : 1) Point charge 2) Line charge 3) Sheet of charge 4) Spherical conducting shell

Answer 1

1)POINT CHARGE :

Electric field is defined as the electric force per unit charge. The electric field of a point charge can be obtained using Coloumb's law. Consider 'q' as the point charge. The electric field is radially outward from the point charge in all directions. The circles formed will behave as equipotential surfaces

E = kq/r²

2)LINE CHARGE:

We derive an expression for the electric field near a line of charge. The result will show the electric field near a line of charge falls off as 1 / a 1/a 1/a , where a is the distance from the line. Assume we have a long line of length L, with total charge Q. Assume the charge is distributed uniformly along the line.

E = λ/2πε₀r   where r is the perpendicular distance from the line charge

3)SHEET OF CHARGE:

For an infinite sheet of charge, the electric field will be perpendicular to the surface. Therefore only the ends of a cylindrical Gaussian surface will contribute to the electric flux . In this case a cylindrical Gaussian surface perpendicular to the charge sheet is used.

E = σ/2ε₀

4)SPHERICAL CONDUCTING SHELL:

Now, the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero.

E = 0

Answer 2

1)POINT CHARGE :

Electric field is defined as the electric force per unit charge. The electric field of a point charge can be obtained using Coloumb's law. Consider 'q' as the point charge. The electric field is radially outward from the point charge in all directions. The circles formed will behave as equipotential surfaces

E = kq/r²

2)LINE CHARGE:

We derive an expression for the electric field near a line of charge. The result will show the electric field near a line of charge falls off as 1 / a 1/a 1/a , where a is the distance from the line. Assume we have a long line of length L, with total charge Q. Assume the charge is distributed uniformly along the line.

E = λ/2πε₀r   where r is the perpendicular distance from the line charge

3)SHEET OF CHARGE:

For an infinite sheet of charge, the electric field will be perpendicular to the surface. Therefore only the ends of a cylindrical Gaussian surface will contribute to the electric flux . In this case a cylindrical Gaussian surface perpendicular to the charge sheet is used.

E = σ/2ε₀

4)SPHERICAL CONDUCTING SHELL:

Now, the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero.

E = 0