Q.A hemispherical shell of uniform mass density is oriented like a bowl, with its flat circular face horizontal and on top and its curved surface bulging downward. Let $P$ be an arbitrary point lying in the plane of the flat circular face (off the central axis). Choose the correct alternative for the direction of the gravitational intensity (gravitational field) at the point $P$.

The figure shows a hemispherical shell of uniform mass density, drawn in cross-section as a bowl shape. Its centre (the centre of the full sphere the hemisphere is half of) is marked C, sitting on the flat rim of the bowl. Three arrows radiate from C in different directions, labelled a (pointing right), b (pointing up), and c (pointing down) -- these are the candidate directions for the gravitational intensity a test mass would feel if placed exactly at C.

A second point, P, is marked at an arbitrary location inside the shell (not at the centre). Four more arrows radiate from P, labelled d, e, f, g, in different directions -- these are the candidate directions for the gravitational intensity at this off-centre point.

This figure grounds Exercises 7.10 and 7.11, which ask the student to reason out, from the shell's symmetry, which single labelled arrow correctly represents the true direction of the gravitational field at C and at P respectively.

Setting up

Complete the hemispherical bowl into a full spherical shell of the same radius and surface density by adding a mirror-image upper hemisphere (the same shape, reflected across the flat face). The point $P$ lies in the plane of the flat face, i.e. on the equatorial plane of this completed sphere — which is inside the completed shell.

Applying the shell theorem

For a complete uniform spherical shell, the gravitational field at every interior point is exactly zero (Newton's shell theorem). Writing $\vec{E}_L$ and $\vec{E}_U$ for the fields at $P$ due to the lower hemisphere (the actual bowl) and the (imagined) upper hemisphere:

Using mirror symmetry to pin the direction

Reflecting across the flat (equatorial) plane maps the upper hemisphere onto the lower one, and maps the point $P$ (which lies exactly in that plane) onto itself. Under this reflection, the horizontal (in-plane) component of a vector stays the same, while the vertical component reverses sign. Applying this to the two hemispheres' fields at $P$: …

Step 1: Complete the bowl into a full spherical shell by adding its mirror-image (upper) hemisphere. By Newton's shell theorem, the field at any point strictly inside a closed uniform shell is zero, so at P: $\vec E_{\text{lower}} + \vec E_{\text{upper}} = 0$.

Step 2: P lies exactly in the mirror plane separating the two hemispheres. Reflecting the whole system across that plane maps the upper hemisphere onto the lower one, and transforms the field vector at P as: horizontal (in-plane) component unchanged, vertical (axial) component reversed.

Step 3: So if $\vec E_{\text{lower}} = (E_h, E_v)$, then $\vec E_{\text{upper}} = (E_h, -E_v)$. Adding: $(2E_h, 0) = 0 \Rightarrow E_h = 0$. …