A vertical cylindrical conducting shell has a charge —q distributed
evenly at either bases with radius r and height h.
Which gaussian surface has an electric field equal to zero?
Vertical cylinder with R > r and H <= h
Vertical cylinder with R =< r and H > h
Vertical cylinder with R > r and H >= h
Vertical cylinder with R =< r and H < h
The electric field due to a uniformly charged cylinder along its axis is given by:
E = (λ / 2πε0) * [R2 / (R2 + L2)^(3/2)]
where λ is the linear charge density (charge per unit length) of the cylinder, R is the distance from the center of the cylinder to the point where the electric field is to be calculated, L is the length of the cylinder, and ε0 is the permittivity of free space.
In this case, the cylinder has a finite height h, and the charge is distributed evenly at the two bases. Therefore, the linear charge density λ = -q / h.
To find the Gaussian surface where the electric field is zero, we need to imagine a cylinder of radius R and height H around the original cylinder. The electric field at any point inside this Gaussian surface will be zero if the enclosed charge is zero.
Let's consider the four options given:
1. Vertical cylinder with R > r and H <= h:
In this case, the Gaussian surface does not enclose any charge. Therefore, the electric field is zero.
2. Vertical cylinder with R =< r and H > h:
In this case, the Gaussian surface encloses the entire charge -q. Therefore, the electric field is not zero.
3. Vertical cylinder with R > r and H >= h:
In this case, the Gaussian surface encloses the entire charge -q. Therefore, the electric field is not zero.
4. Vertical cylinder with R =< r and H < h:
In this case, the Gaussian surface encloses only a portion of the charge -q. Therefore, the enclosed charge is not zero, and the electric field is not zero.
Therefore, the correct answer is option 1: Vertical cylinder with R > r and H <= h.