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# 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 directedA. perpendicular to the diameterB. parallel to the diameterC. at an angle tilted towards the diameterD. at an angle away from the diameter

Last updated date: 18th Jun 2024
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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.

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.