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

NCERT Exemplar · Q7

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

(a) perpendicular to the diameter

(b) parallel to the diameter

(d) at an angle tilted away from the diameter.

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Concept understanding — Gauss Law

Gauss's Law is a shortcut. Instead of adding up the Coulomb contribution of every charge — a nightmare of vectors and integrals — it lets you read the field straight off the symmetry of a problem. The whole idea rests on one quantity: electric flux.

Flux — field-lines counted through a surface. For a uniform field E crossing a flat area A, the flux is Φ = E·A = EA cosθ, where θ is the angle between the field and the normal to the surface. Picture the field as a bundle of lines; flux is how many pierce the surface. When E is edge-on (θ = 90°) nothing passes through and Φ = 0; when E is ⊥ to the surface (θ = 0) the count is maximal. For a closed surface, Φ is the net number of lines leaving it — lines that enter and exit cancel.

The law itself. Gauss's law states that the net flux out of any closed surface depends only on the charge trapped inside:

Φ = q_enclosed / ε₀.

Two consequences do most of the work. First, only enclosed charge counts — a charge outside sends as many lines in as out, so its net contribution is exactly zero. Second, the surface's shape is irrelevant; move the charge around inside or deform the surface, and Φ never changes.

Why symmetry makes it powerful. By itself Φ = q/ε₀ has E buried in an integral. It becomes a tool only when you pick a Gaussian surface matched to the symmetry — one where E is constant and everywhere either ⊥ to the surface (so Φ = EA) or ∥ to it (contributing nothing). Then E slides out and you solve in one line. This works for exactly three geometries:

1 — Infinite line charge (linear density λ). Use a coaxial cylinder: E = λ / 2πε₀r, falling off as ∝ 1/r.

2 — Infinite sheet (surface density σ). Use a pillbox pierced through the sheet: E = σ / 2ε₀ — uniform and completely independent of distance. The field near a large charged plane simply doesn't weaken as you step back. …

Why this formula?

Gauss's Law: Why It Holds

Gauss's Law is one of the four Maxwell's equations and a cornerstone of electromagnetism. Let's build the understanding from the ground up — not just the formula, but the why.

1. The Core Idea: Flux as "Flow" of Field

Imagine an electric field $\vec{E}$ passing through a small patch of area $dA$ . The electric flux through that patch is:

d\Phi_E = \vec{E} \cdot d\vec{A} = E \, dA \cos\theta

where $\theta$ is the angle between $\vec{E}$ and the outward normal to the surface.

Why this definition?

If $\vec{E}$ is perpendicular to the surface ( $\theta = 0$ ), maximum field "flows through".
If $\vec{E}$ is parallel ( $\theta = 90^\circ$ ), no flux — the field just slides along the surface.

Total flux through a closed surface $S$ is:

\Phi_E = \oint_S \vec{E} \cdot d\vec{A}

2. The Key Insight: Flux Depends Only on Enclosed Charge

Consider a single point charge $+q$ at the centre of a spherical surface of radius $r$ .

By Coulomb's law, at every point on the sphere: $E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$ , radially outward.
The area vector $d\vec{A}$ is also radially outward.
So $\vec{E} \cdot d\vec{A} = E \, dA$ everywhere.

The total flux:

\Phi_E = \oint E \, dA = E \oint dA = \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \right) (4\pi r^2) = \frac{q}{\varepsilon_0}

Notice: The $r^2$ cancels! The flux is independent of the radius.

3. Why Shape Doesn't Matter

Now, what if the surface is not spherical but any closed shape enclosing the charge?

Draw a small cone from the charge to the surface.
The flux through a patch $dA$ at distance $r$ is $d\Phi = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} \cos\theta \, dA$ .
But $\frac{\cos\theta \, dA}{r^2}$ is exactly the solid angle $d\Omega$ subtended by $dA$ at the charge.

So $d\Phi = \frac{q}{4\pi\varepsilon_0} d\Omega$ .

Integrating over the entire closed surface: $\oint d\Omega = 4\pi$ (total solid angle around a point).

Hence:

\Phi_E = \frac{q}{4\pi\varepsilon_0} \cdot 4\pi = \frac{q}{\varepsilon_0}

Result: For any closed surface enclosing $q$ , the flux is $\frac{q}{\varepsilon_0}$ .

4. Multiple Charges: Superposition

If there are many charges $q_1, q_2, \dots, q_n$ inside the surface, the total electric field is the vector sum of individual fields:

\vec{E} = \vec{E}_1 + \vec{E}_2 + \dots + \vec{E}_n

Flux is linear: …

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By mirror symmetry the field at $P$ lies in the plane containing the axis and the diameter; it is purely axial only at the centre and becomes increasingly aligned with the diameter near the rim, so at a general point away from the centre it is tilted towards the diameter — option (c).

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. …

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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:

Choose a coordinate system (e.g., axis along the diameter).
For each charge element, find the field direction.
Cancel components perpendicular to the axis (they sum to zero).
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. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 33 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$Q. A spherical rubber balloon has the electric charges uniformly distributed over its surface. If the balloon is inflated further, the electric intensity E on the surface (A) increases (B) decreases (C) remains same (D) is zero$
›Reveal solutionSolution
This tests how surface field strength changes for a charged sphere whose radius grows while total charge stays constant; the field decreases.

Concept and Intuition

Inflating the balloon spreads the same total charge $Q$ over a larger surface area, so the surface charge density $\sigma = Q/4\pi R^2$ drops. The electric field just outside a uniformly charged sphere depends only on $Q$ and $R$ (by Gauss's law, exactly as if all charge were concentrated at the centre), so as $R$ grows for fixed $Q$ , the field at the surface must weaken.

Step-by-Step Solution

By Gauss's law, the field at the surface of a uniformly charged sphere of radius $R$ carrying charge $Q$ is $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$ .

Inflating the balloon increases $R$ but does not add or remove any charge, so $Q$ stays constant. …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$70\hat i$
›Reveal solutionSolution
A straightforward flux calculation: since the uniform field is parallel to the cylinder's axis, the flux through each flat end face equals $E$ times the face's circular area, giving a magnitude of about 55 (SI units).

Concept and Intuition

Electric flux through a flat surface in a uniform field is $\Phi = \vec E \cdot \vec A = EA\cos\theta$ , where $\theta$ is the angle between the field and the surface's (outward) normal. For a cylinder with its axis along $x$ and a uniform field $\vec E = 70\hat i$ also along $x$ , the two flat circular end faces have their normals exactly along (or exactly against) the field direction — so $\cos\theta = \pm1$ for these faces, and the flux magnitude reduces to simply $E$ times the face area. (The curved lateral surface, by contrast, has its normal everywhere perpendicular to the axis, so it carries zero flux — consistent with the field lines simply passing straight through the cylinder without net flux crossing the sides.)

Step-by-Step Solution

The left face is a circular disk of radius $r=0.5\ m$ , with area $A=\pi r^2$ .

Compute the area: $A = \pi\times(0.5)^2 = \pi\times0.25 \approx \dfrac{22}{7}\times0.25 = 0.7857\ m^2$ . …
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AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$720\ NC^{-1}$
›Reveal solutionSolution
This tests the field between oppositely-charged parallel plates, $E=\sigma/\varepsilon_0$ , to find the charge on each plate. Answer: 5 nC.

Concept and Intuition

For a single charged plate (infinite sheet), the field on each side is $\sigma/2\varepsilon_0$ . But for two large plates with equal and opposite charges facing each other (a parallel-plate-capacitor-like arrangement), the fields from both plates point in the same direction in the region between them, so they add: $E_{between} = \dfrac{\sigma}{2\varepsilon_0}+\dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0}$ .

Step-by-Step Solution

Field between two oppositely charged plates: $E = \dfrac{\sigma}{\varepsilon_0}$ .

Solve for surface charge density: $\sigma = E\varepsilon_0 = 720 \times 8.854\times10^{-12} = 6.3749\times10^{-9}\ C/m^2$ .

Area of each circular plate: $A = \pi r^2 = \pi (0.5)^2 = \pi(0.25) \approx 0.7854\ m^2$ . …
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AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$q/a^2$
›Reveal solutionSolution
Tests the core statement of Gauss's law: flux through a closed surface depends only on enclosed charge, not on the surface's shape, size, or the charge's exact position inside.

Concept and Intuition

Gauss's law is a global statement — it doesn't care whether the enclosing surface is a sphere, a cube, or any irregular shape, and it doesn't care exactly where inside the charge sits. All that matters is how much charge is inside. This is precisely why Gauss's law is so powerful for symmetric charge distributions: you're free to choose whatever surface makes the calculation easiest.

Step-by-Step Solution

Gauss's law states $\oint \vec E\cdot d\vec A = \dfrac{q_{enc}}{\epsilon_0}$ for any closed surface.

Here the closed surface is a cube of side $a$ that encloses a charge $q$ .

Regardless of the cube's side length $a$ or the position of $q$ inside it, the total flux is simply $\Phi = \dfrac{q}{\epsilon_0}$ . …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$8.0\times10^3\ \mathrm{Nm^2C^{-1}}$
›Reveal solutionSolution
Gauss's law directly relates the flux through a closed surface to the enclosed charge: $q=\varepsilon_0\Phi\approx70\,nC$ .

Concept and Intuition

Gauss's law states that the net electric flux through any closed surface depends only on the charge enclosed, regardless of the shape of the surface or the position of the charge inside it: $\Phi=\dfrac{q_{enc}}{\varepsilon_0}$ .

Step-by-Step Solution

Given $\Phi=8.0\times10^3\,\mathrm{Nm^2C^{-1}}$ .

Rearranging Gauss's law: $q_{enc}=\varepsilon_0\Phi$ .

Substitute $\varepsilon_0=8.85\times10^{-12}\,\mathrm{C^2N^{-1}m^{-2}}$ : $q_{enc}=8.85\times10^{-12}\times8.0\times10^3=7.08\times10^{-8}\,C$ .

Convert: $7.08\times10^{-8}\,C=70.8\,nC\approx70\,nC$ . …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$\dfrac{3\epsilon_0}{R^2}$
›Reveal solutionSolution
This tests Gauss's law applied to a uniformly charged solid sphere — the field just outside the surface behaves as if all charge were concentrated at the centre.

Concept and Intuition

For any spherically symmetric charge distribution, Gauss's law lets us treat the enclosed charge as if it were a point charge at the centre when evaluating the field outside (or at) the surface. So we first find the total charge in terms of the given volume charge density, then apply the point-charge field formula at $r=R$ .

Step-by-Step Solution

Total charge enclosed by the sphere: $Q=\rho \times \text{Volume} = \rho\cdot\dfrac{4}{3}\pi R^3$ .

By Gauss's law, the field at the surface ( $r=R$ ) is $E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R^2}$ . …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$\dfrac{Q}{3}$
›Reveal solutionSolution
Grounding the inner shell of a concentric-shell system forces its potential to zero; solving for the resulting charge shows exactly $Q/2$ flows to earth.

Concept and Intuition

When a conductor is grounded, its potential is fixed at zero (earth's potential), and charge flows onto or off it until that condition is met. For concentric shells, we use the fact that inside a uniformly charged shell the potential is constant and equals $kq_{shell}/R_{shell}$ , and we superpose contributions from every shell of charge present.

Step-by-Step Solution

Let the final charge on the inner shell (radius $r$ ) be $q$ . The outer shell (radius $2r$ ) has total charge $Q$ ; by electrostatic induction its inner surface carries $-q$ and its outer surface carries $Q+q$ .

Potential of the inner shell = (its own contribution) + (contribution of outer shell's charges, since inner shell is inside the outer shell):

$V = \frac{kq}{r} + \frac{k(-q)}{2r} + \frac{k(Q+q)}{2r} = \frac{kq}{r} + \frac{kQ}{2r}$

Setting $V=0$ (grounded): $\dfrac{q}{r} = -\dfrac{Q}{2r} \Rightarrow q = -\dfrac{Q}{2}$ . …
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AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$Q. Two charged conducting spheres of radii 5 cm and 10 cm have equal surface charge densities. If the electric field on the surface of the smaller sphere is E, then the electric field on the surface of the larger sphere is (A) 2E (B) 4E (C) 0.5E (D) E$
›Reveal solutionSolution
This tests a subtlety of conductors: the field just outside a charged conducting surface depends only on the local surface charge density, not on the sphere's radius. Equal $\sigma$ means equal E, regardless of size — the answer is E (unchanged).

Concept and Intuition

It's tempting to reach for $E = kQ/r^2$ and think a bigger sphere (different $Q$ , different $r$ ) must have a different field. But the correct starting point for a conductor's surface is Gauss's law applied to a small pillbox straddling the surface, which gives the general boundary condition $E = \sigma/\varepsilon_0$ for ANY conductor surface — this holds locally, independent of the conductor's overall size or shape. Since surface charge density is defined as $\sigma = Q/A = Q/(4\pi r^2)$ , a sphere's total charge Q does scale with $r^2$ for a given $\sigma$ , but the field expressed in terms of $\sigma$ directly doesn't care about $r$ at all.

Step-by-Step Solution

Field just outside any conductor's surface: $E = \dfrac{\sigma}{\varepsilon_0}$ (a standard result from Gauss's law at a conducting boundary).

This expression contains no explicit dependence on the sphere's radius.

Both spheres (radii 5 cm and 10 cm) are given to have the SAME surface charge density $\sigma$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\bar{E} = (3\hat{i} + 5\hat{j} + 7\hat{k})\ \mathrm{N\,C^{-1}}$
›Reveal solutionSolution
Electric flux through a surface only depends on the field component perpendicular to it; for the yz-plane that's $E_x$ , giving $\Phi=9$ (SI units).

Concept and Intuition

Electric flux is the dot product $\Phi = \vec E\cdot\vec A$ , where $\vec A$ points along the surface's normal. A surface lying in the yz-plane has its normal along the x-axis, so components of $\vec E$ along $\hat j$ and $\hat k$ lie within the plane and contribute nothing to flux through it — only the $\hat i$ component matters.

Step-by-Step Solution

Surface lies in the yz-plane $\Rightarrow$ its area vector is $\vec A = 3\hat i\ \mathrm{m^2}$ (magnitude 3, along x).

Given $\vec E = (3\hat i+5\hat j+7\hat k)\ \mathrm{N/C}$ . …
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AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$\mu$
›Reveal solutionSolution
Gauss's law says the flux through a closed surface depends only on the enclosed charge, never on the surface's size or shape — giving $72\pi\times10^3\ \text{N m}^2\text{C}^{-1}$ regardless of the 6 cm radius given.

Concept and Intuition

Gauss's law states $\oint \vec E\cdot d\vec A = \dfrac{q_{enc}}{\varepsilon_0}$ . This is a purely topological statement — as long as the same charge is enclosed, the total flux out of any closed surface around it (sphere, cube, irregular blob) is identical. The radius of the shell given here is a distractor; it doesn't enter the flux calculation at all.

Step-by-Step Solution

Enclosed charge: $q=2\ \mu\text{C} = 2\times10^{-6}$ C.

$\Phi = \dfrac{q}{\varepsilon_0}$ . Using $\dfrac1{4\pi\varepsilon_0}=k=9\times10^9$ , we get $\dfrac1{\varepsilon_0}=4\pi k$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$\dfrac{4Q}{\varepsilon_0}$
›Reveal solutionSolution
Gauss's law says the flux through a closed surface depends only on the enclosed charge, not on the surface's size or shape; since the imaginary sphere of radius 2R encloses the same charge Q as the inner sphere, the flux is simply $Q/\varepsilon_0$ .

Concept and Intuition

Gauss's law, $\oint \vec E \cdot d\vec A = \dfrac{Q_{enc}}{\varepsilon_0}$ , is a statement about total enclosed charge, completely independent of the geometry of the enclosing (Gaussian) surface or the exact charge distribution inside it. Doubling or tripling the radius of the imaginary sphere changes nothing about the flux, as long as no additional charge is included or excluded.

Step-by-Step Solution

The imaginary sphere of radius $2R$ is concentric with, and completely encloses, the charged sphere of radius $R$ and charge $Q$ .

Total charge enclosed by the imaginary sphere: $Q_{enc} = Q$ (no additional charge outside the inner sphere but inside the imaginary sphere). …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$\pm 6 \times 10^{-6}$
›Reveal solutionSolution
Gauss’s law says the net electric flux through a closed surface equals the net charge enclosed divided by ε₀. Since a dipole has equal and opposite charges, the net enclosed charge is zero, so the total flux is zero.

The key insight is that Gauss’s law relates the total electric flux through a closed surface to the net charge inside that surface — not to the arrangement or separation of the charges. A dipole consists of two equal-magnitude, opposite-sign charges. When both are fully inside the closed surface, their contributions to the net enclosed charge cancel exactly.

State Gauss’s law For any closed surface, the total electric flux Φₑ is given by

$\Phi_e = \frac{Q_{\text{enc}}}{\varepsilon_0}$
where $Q_{\text{enc}}$ is the algebraic sum of all charges inside the surface.

Identify the enclosed charges The dipole has $+Q = +6 \times 10^{-6}\ \text{C}$ at one end and $-Q = -6 \times 10^{-6}\ \text{C}$ at the other. Both charges lie completely inside the oval surface.

$Q_{\text{enc}} = (+6 \times 10^{-6}) + (-6 \times 10^{-6}) = 0$

Apply Gauss’s law With $Q_{\text{enc}} = 0$ ,

$\Phi_e = \frac{0}{\varepsilon_0} = 0$
The flux is zero regardless of the shape of the surface or the positions of the charges inside, as long as both are enclosed. …
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Q.A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed

Concept understanding — Gauss Law

Why this formula?

Gauss's Law: Why It Holds

1. The Core Idea: Flux as "Flow" of Field

2. The Key Insight: Flux Depends Only on Enclosed Charge

3. Why Shape Doesn't Matter

4. Multiple Charges: Superposition

Setting up the symmetry

Concept: Superposition & Symmetry in Electrostatics

Method: Superposition of Two Half-Spheres

Why this method?

Steps

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

Mistake 2: Thinking the field points toward the flat face

Mistake 3: Forgetting to use vector addition (superposition)

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

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

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

Setting up the symmetry

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