A nonconducting hemispherical cup of inner radius R has a total charge q spread uniformly over its inner surface. We have to find the electric field at the centre of curvature.

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We are given a nonconducting hemispherical cup of inner radius R that has a total charge q spread uniformly over its inner surface. We will calculate the electric field at the centre of curvature of the hemisphere, which is at the point P, in two steps.

In the first step we will find the electric field at a point y unit above the centre of a ring of radius r and on the axis passing through its centre.

Let the charge on the ring be . From the symmetry of the charge distribution we note that the electric field at P will be along the y-axis and its magnitude will be

In the second step we will consider the hemispherical cup as a stack of rings, as shown in the figure. Let us consider a ring at angle from the azimuth, as shown in the figure.

Perpendicular distance of the ring from the centre of curvature of the hemispherical cup is . Area of this ring on the surface of the hemispherical cup will be

As the amount of charge on the inner surface of the hemispherical cup is q, the surface charge density is

.

Charge on the ring, .

Using the result of step 1, we have

By integrating from , we calculate the electric field at the centre of curvature of the hemispherical cup. We have

.

Substituting for , we find

. Its direction is along the perpendicular axis passing through the centre of curvature of the hemisphere.