Question

A hemisphere shell is uniformly charged positively. The electric field at a point on a diameter away from the centre (inside the boundary of hemispherical shell) is directed

A. perpendicular to the diameter

B. parallel to the diameter

C. at an angle tilted towards the diameter

D. at an angle away from the diameter

Answer 1

Hint: Go through the basics of the electric field intensity at a point due to positive charge on the surface. Show the direction of the electric field at the point on the diameter away from the centre and the direction of its horizontal and vertical components. Hence, determine the direction of the net electric field at that point.

Complete step by step answer:

We have given that a hemispherical shell is uniformly charged with positive charge. We are asked to determine the direction of the electric field at a point on the diameter away from the centre. As the hemispherical shell is positively charged, the positive charge is present only on the outer surface of the hemisphere as the inner space of the hemisphere is hollow and empty.

Let us consider a test charge placed at a point P on the diameter of the hemispherical shell inside the boundary of the hemispherical shell.

Let us consider the electric field at point P due to the positive charges on the two edges of the diameter of the hemispherical shell. The direction of the electric field \[E\] at point P due to the positive charges on the edges of diameter of the hemispherical shell is as shown in the figure. The horizontal and vertical components of the electric field at point P are also shown in the figure.

From the diagram, we can conclude that the horizontal components of the electric field \[{E_x}\] at point P are parallel to the diameter of the hemispherical shell and directed opposite to each other. The vertical components of the electric field \[{E_y}\] at point P are in the direction perpendicular to the diameter and both the vertical components combine together to give the net electric field at point P on the diameter of the hemispherical shell. Therefore, the direction of the electric field at a point on the diameter away from the centre is perpendicular to the electric field.

Hence, the correct option is A.

Note: The students may think that how can the horizontal components of the electric field at point P cancel each other as we don’t know the magnitude of these components. But we have assumed the electric field due to the same positive charge on the edges of diameter at a point P on the diameter at the same distance from these two positive charges. Hence, the horizontal and vertical components of the electric field at point P should have the same magnitude.