Q.A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed

Concept: field of a uniformly charged hemisphere at points on its base.

By mirror symmetry about the plane through the axis and the diameter, the field at any point $P$ on the diameter has no component out of that plane. At the centre, full symmetry makes the field purely axial (perpendicular to every diameter). Moving to a point away from the centre, the nearby part of the curved shell (whose outward normal there is nearly along the diameter) contributes an increasingly la …

Setting up the symmetry

Model the hemisphere as a uniformly (positively) charged hemispherical shell of radius $R$, flat circular face in a plane, with a diameter of that face lying along, say, the $x$-axis through the centre $O$. Let $P$ be a point on this diameter at distance $d$ from $O$ ($0<d<R$), still in the plane of the flat face.

Step 1 — Kill the out-of-plane component. The hemisphere is symmetric under reflection through the plane containing the axis (the $z$-axis, perpendicular to the base) and the chosen diameter (the $xz$-plane). Every charge element at $y>0$ has a mirror partner at $-y$ contributing an equal and opposite $y$-component of field at $P$ (which itself sits at $y=0$). So $E_y(P)=0$: the resultant field must lie in the $xz$-plane, i.e., in the plane of the axis and the diameter.

Step 2 — What happens exactly at the centre. At $O$ ($d=0$), the hemisphere has full rotational symmetry about the axis, so by the same mirror argument applied to EVERY diameter through $O$, all horizontal components cancel and only the axial ($z$) component survives. This is the familiar result $E_{O}=\dfrac{\sigma}{4\varepsilon_0}$, directed along the axis, away from the curved surface — i.e., perpendicular to every diameter.

Step 3 — What happens near the rim. As $P$ moves out to $d\to R$ (near the edge of the flat face), it approaches the ring where the curved surface meets the base — the "equator." Right there, the nearby patch of the curved shell is almost tangent to a vertical cylinder, i.e., its outward normal is nearly horizontal, along the diameter direction itself. A point just inside that patch sits in the field of what looks locally like a charged sheet whose normal is along the diameter — so the dominant, nearby contribution to $E$ at points close to the rim is along the diameter, not axial. …

Concept: Superposition & Symmetry in Electrostatics

For a uniformly charged hemisphere, the electric field at a point on the axis (the diameter line) is not simply radial — it has a net direction due to the broken spherical symmetry.

Method: Superposition of Two Half-Spheres

Why this method?

A full uniformly charged sphere produces zero net field at its centre (by symmetry). A hemisphere is exactly half of that sphere. So we can think:

Full sphere = Hemisphere A + Hemisphere B (identical, oppositely oriented)

At the centre of the full sphere, the field is zero. Therefore, the field due to one hemisphere must be equal in magnitude and opposite in direction to the field due to the other hemisphere.

Steps

1. Imagine a full sphere of radius $R$, uniformly charged with total charge $+2Q$ (so each hemisphere has charge $+Q$).

2. At the centre O of the full sphere, by symmetry:

3. Let $\vec{E}_{\text{hemi}}$ be the field at O due to one hemisphere (say the upper half).

4. The other hemisphere (lower half) produces field $-\vec{E}_{\text{hemi}}$ at O, so that:

5. Key result: The field at the centre of a uniformly charged hemisphere is not zero — it points away from the flat face (if positively charged). …

Here are the common mistakes students make when analyzing the electric field direction for a uniformly charged hemisphere, along with how to avoid each.

Mistake 1: Assuming the field is radial (like a full sphere)

The error: Students treat the hemisphere like a full sphere and conclude the field at a point on the axis (the diameter) is directed radially outward (away from the centre in all directions).

Why it’s wrong: A full sphere has spherical symmetry — every bit of charge pulls equally in all directions, so the net field at the centre is zero, and outside it is radial. A hemisphere breaks that symmetry. There is no charge on the missing half, so the field cannot be purely radial.

How to avoid: Always check for symmetry first.

• Full sphere: Symmetric → radial field.
• Hemisphere: Only half the charge exists. The missing half means the field will point away from the centre but also away from the flat face (i.e., along the axis, away from the flat side).

Mistake 2: Thinking the field points toward the flat face

The error: Some students imagine the field lines “leaking” out of the flat circular face and conclude the field points toward that face.

Why it’s wrong: The hemisphere is positively charged. Electric field lines point away from positive charge. The flat face has no charge (it’s an imaginary surface), so the field cannot point toward it. The field must point away from the bulk of the positive charge.

How to avoid: Remember the fundamental rule:

• Positive charge → field lines radiate outward.
• The field at any point is the vector sum of contributions from all charge elements. For a point on the axis (the diameter), the net field points away from the centre and away from the flat face — i.e., along the axis, outward from the curved side.

Mistake 3: Forgetting to use vector addition (superposition)

The error: Students try to guess the direction intuitively without summing contributions from all parts of the hemisphere.

Why it’s wrong: The field at a point is the vector sum of fields from every infinitesimal charge element. Without superposition, you miss that horizontal components cancel (due to symmetry about the axis) but vertical components add.

How to avoid: Always break the problem into components:

1. Choose a coordinate system (e.g., axis along the diameter).
2. For each charge element, find the field direction.
3. Cancel components perpendicular to the axis (they sum to zero).
4. Add components along the axis — they all point in the same direction (away from the flat face).

Mistake 4: Confusing “on a diameter” with “at the centre”

The error: Students think “on a diameter” means exactly at the centre of the hemisphere.

Why it’s wrong: The question says “on a diameter away from the centre” — meaning a point on the axis outside the centre, not at it. At the exact centre, the field is not zero (unlike a full sphere), but the direction is still along the axis away from the flat face.

How to avoid: Read carefully:

• “On a diameter” = on the axis of symmetry.
• “Away from the centre” = not at the centre, but somewhere along that line. …