Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region, (a) the electric field is necessarily zero. (b) the electric field is due to the dipole moment of the charge distribution only. (c) the dominant electric field is $\propto \dfrac{1}{r^3}$, for large r, where r is the distance from a origin in this region. (d) the work done to move a charged particle along a closed path, away from the region, will be zero.

Consider a region inside which there are various types of charges but the total charge is zero. At points outs

NCERT Exemplar · Q11

Q. Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region,

(a) the electric field is necessarily zero.

(b) the electric field is due to the dipole moment of the charge distribution only.

\propto \dfrac{1}{r^3}

, for large r, where r is the distance from a origin in this region.

(d) the work done to move a charged particle along a closed path, away from the region, will be zero.

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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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With zero total charge inside, the monopole term of the field vanishes; the leading surviving term at large distance is the dipole term, which falls off as $1/r^3$ — option (c) — and since the electrostatic field is always conservative, the work done around ANY closed path is zero regardless of the charge configuration — option (d). Both (c) and (d) are correct; (a) and (b) are false.

Multipole expansion: what "zero net charge" does and doesn't tell you

Any finite charge distribution can be described, from far away, by an expansion in powers of $1/r$ :

V(\vec r) = \frac{1}{4\pi\varepsilon_0}\left[\frac{Q}{r} + \frac{\vec p\cdot\hat r}{r^2} + \cdots\right],

where $Q$ is the total (monopole) charge and $\vec p$ is the dipole moment of the distribution. The corresponding field terms fall off one power of $r$ faster than the potential terms: the monopole field goes as $1/r^2$ , the dipole field as $1/r^3$ , and so on.

Step 1 — Kill the monopole term. We are told the total charge in the region is zero, $Q=0$ . So the $1/r$ potential term (and its $1/r^2$ field) vanishes identically — there is no ordinary Coulomb-type field surviving at large $r$ .

Step 2 — Identify the next surviving term. With $Q=0$ , the next term in the expansion is the dipole term, $\vec p\cdot\hat r/r^2$ in potential, giving a field that falls as $1/r^3$ . Unless the dipole moment $\vec p$ also happens to be zero (a special, non-generic case), this dipole term dominates at large $r$ — it is the leading surviving contribution. This is exactly option (c): the dominant field for large $r$ is $\propto 1/r^3$ .

Step 3 — Check option (a). "The electric field is necessarily zero" is false — the dipole (or higher multipole) term is in general non-zero outside the region, exactly as in Step 2.

Step 4 — Check option (b). "The electric field is due to the dipole moment only" over-claims: the dipole term is only the dominant (leading) term for large $r$ ; there can also be quadrupole and higher contributions (smaller, falling as $1/r^4$ , etc.), and if $\vec p$ itself happens to vanish, the field would be dominated by the quadrupole term instead. So the field is not purely dipolar in general — (b) is false as a blanket statement. …

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Here is a breakdown of the common mistakes students make when analyzing the electric field and potential outside a region with zero net charge, along with how to avoid them.

The Core Concept: Net Charge vs. Charge Distribution

The key idea is that zero total charge does not imply zero electric field or zero potential outside the region. The field depends on how the charge is arranged (the dipole moment, quadrupole moment, etc.), not just the total sum.

Mistake 1: Assuming the Electric Field is Zero Everywhere Outside

The Error:

Students often think: "Total charge = 0, so by Gauss's Law, the flux through any closed surface outside the region is zero, therefore the field must be zero."

Why it's wrong:

Gauss's Law says $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$ . If $Q_{\text{enc}} = 0$ , the net flux is zero, but the field itself can be non-zero. For example, a dipole (equal +q and -q separated by a small distance) has zero net charge, but outside the dipole, the field is clearly non-zero.

How to Avoid:

Remember: Flux = 0 does not mean Field = 0. It only means the net number of field lines entering equals the number leaving.
Visualize a dipole: field lines loop from +q to -q. Outside, they still exist.
Key exam fact: For a region with zero net charge, the leading term of the field at large distances is the dipole field ( $\propto 1/r^3$ ), not zero.

Mistake 2: Assuming the Electric Potential is Zero Everywhere Outside

The Error:

Students think: "If total charge is zero, the potential at infinity is zero, so the potential everywhere outside must also be zero."

Why it's wrong:

Potential is a scalar sum: $V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \sum \frac{q_i}{|\vec{r} - \vec{r}_i|}$ . Even if $\sum q_i = 0$ , the distances $|\vec{r} - \vec{r}_i|$ are different for each charge, so the contributions don't cancel exactly. For a dipole, $V \propto \frac{\cos\theta}{r^2}$ , which is non-zero.

How to Avoid:

Always write the potential as a sum of contributions from individual charges. Do not cancel charges algebraically before considering distances.
Rule of thumb: If the charges are not at the same point, the potential outside is generally non-zero.
For a spherically symmetric charge distribution with zero net charge, the potential is zero outside (because the field is zero by Gauss's Law). But for any non-symmetric arrangement, it is not.

Mistake 3: Confusing "Total Charge Zero" with "No Charges Present"

The Error:

Students treat the region as if it were empty space, concluding $\vec{E}=0$ and $V=0$ everywhere.

Why it's wrong:

The problem explicitly says "various types of charges" are present, just that they sum to zero. This is a neutral system (like an atom or a dipole), not a vacuum.

How to Avoid:

Read carefully: "total charge is zero" ≠ "no charge".
Think of a neutral atom: equal protons and electrons. Outside the atom, there is a non-zero electric field (due to the dipole moment) and potential.
Exam tip: If the question says "various types of charges but total charge zero," immediately think of multipole moments (dipole, quadrupole, etc.).

Mistake 4: Forgetting the Role of Distance and Symmetry

The Error:

Students assume that if the region is small, the field outside is negligible or zero.

Why it's wrong:

Even a tiny dipole (like a water molecule) produces a measurable field at large distances. The field decays as $1/r^3$ (dipole) or faster, but it is not zero.

How to Avoid:

Remember the multipole expansion:
- Monopole term (total charge) = 0 → next term is dipole.
- Dipole term $\propto 1/r^3$ (for field) or $1/r^2$ (for potential).
- If dipole moment is also zero, then quadrupole term $\propto 1/r^4$ appears, etc.
Key formula: For a dipole with moment $\vec{p}$ ,

V(\vec{r}) = \frac{1}{4\pi\epsilon_0} \frac{\vec{p} \cdot \hat{r}}{r^2}

This is clearly non-zero for most directions.

--- …

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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 16 on this concept.

COMEDK 2026Set 2026-A1 markMCQ
$5 \mu \mathrm{C}$
›Reveal solutionSolution
The electric flux through any closed surface depends only on the net charge enclosed, not on the size or shape of the surface. Since both spherical shells enclose the same charge, the flux through each is identical, so the ratio is 1:1.

Concept and Intuition

This problem is a direct application of Gauss’s law:
$\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}$
The flux through a closed surface depends only on the total charge inside it. The radius of the shell is irrelevant — a larger shell just spreads the same flux over a larger area, but the total flux remains unchanged. Many students mistakenly think flux changes with radius, but Gauss’s law says otherwise.

Step-by-step reasoning

Identify the charge enclosed by $S_1$ The inner shell $S_1$ has radius 10 cm and contains the $5\,\mu\text{C}$ charge at its centre. No other charge is inside $S_1$ .

$Q_{\text{enc},1} = 5\,\mu\text{C}$

Flux through $S_1$ By Gauss’s law:

$\Phi_1 = \frac{Q_{\text{enc},1}}{\varepsilon_0} = \frac{5\times10^{-6}}{\varepsilon_0}$

Identify the charge enclosed by $S_2$ The outer shell $S_2$ of radius 20 cm encloses the entire inner system — the charge at the centre and the inner shell $S_1$ (which is neutral, just a conductor). The only net charge inside $S_2$ is still the same $5\,\mu\text{C}$ .

$Q_{\text{enc},2} = 5\,\mu\text{C}$

Flux through $S_2$ Again by Gauss’s law: $\Phi_2 = \frac{Q_{\text{enc},2}}{\varepsilon_0} = \frac{5\times10^{-6}}{\varepsilon_0}$ …
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COMEDK 2026Set 2026-A1 markMCQ
$m$
›Reveal solutionSolution

The field of a charged conducting sheet is $E=\sigma/\varepsilon_0$ , so the horizontal electric force $QE$ makes $\tan\theta\propto\sigma$ — option (A).

The pith ball sits in equilibrium under three forces: its weight $mg$ (vertically down), the tension $T$ along the thread, and the electrostatic force $F$ pushing it horizontally away from the sheet.

The electric field just outside a large charged conducting sheet of surface density $\sigma$ is
$E=\frac{\sigma}{\varepsilon_0}.$
So the horizontal force on the charge is
$F=QE=\frac{Q\sigma}{\varepsilon_0}.$
Resolving the equilibrium of the ball, the thread's deflection satisfies …
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COMEDK 2026Set 2026-M1 markMCQ
$E=3 \hat{i}+6 \hat{j}+\hat{k}$
›Reveal solutionSolution
The key idea is that the total electric flux through a closed surface equals the enclosed charge divided by ε₀ (Gauss’s law). Since flux through five faces is zero, the flux through the given face is the total flux. Compute that flux via dot product of E and the area vector, then solve for charge.

Concept & Intuition

Gauss’s law states that the net electric flux through any closed surface equals the charge enclosed divided by ε₀. Here, the cuboid is a closed surface. We are told the flux through five of its six faces is zero, so the only contribution to the total flux comes from the one face whose area and outward normal are given. Therefore, the flux through that face is the total flux. We compute it as the dot product of the electric field with the area vector (area times outward unit normal). Then set that equal to $Q_{\text{enc}}/\varepsilon_0$ and solve.

Step-by-step solution

Identify the area vector The face has area $A = 4\ \text{m}^2$ and outward unit normal

$\hat{n} = \frac{2\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{17}}.$
The area vector is
$\vec{A} = A\,\hat{n} = 4 \cdot \frac{2\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{17}} = \frac{8\hat{i}+8\hat{j}+12\hat{k}}{\sqrt{17}}.$

Compute the electric flux through this face The electric field is $\vec{E} = 3\hat{i}+6\hat{j}+\hat{k}$ . Flux through a surface is

$\Phi = \vec{E} \cdot \vec{A}.$
So
$\Phi = \frac{1}{\sqrt{17}} \bigl(3\cdot8 + 6\cdot8 + 1\cdot12\bigr) = \frac{1}{\sqrt{17}} (24 + 48 + 12) = \frac{84}{\sqrt{17}}.$

Apply Gauss’s law The total flux through the closed cuboidal surface is
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KCET 2026Set C21 markMCQ
$-2Q, Q$
(a) Net flux through $S_1$
(b) Net flux through $S_2$
(c) Net flux through $S_3$ List-II
(i) $\frac{-2Q}{\epsilon_0}$
(ii) $\frac{-Q}{\epsilon_0}$
(iii) Zero (A) a - ii, b - i, c - iii (B) a - iii, b - ii, c - i (C) a - i, b - ii, c - iii (D) a - ii, b - iii, c - i
›Reveal solutionSolution
Gauss's law states that the net electric flux through a closed surface equals $Q_{enc}/\epsilon_0$ , independent of the position of the charges inside or any charge outside the surface.

Step 1 — Identify the enclosed charge for each surface

From the figure, the three point charges $-2Q$ , $Q$ and $-Q$ are arranged in a line, with $S_1$ , $S_2$ and $S_3$ enclosing different combinations of them:

$S_1$ encloses $-2Q$ and $Q$ , so $Q_{enc,1} = -2Q + Q = -Q$ .

$S_2$ encloses only $Q$ and $-Q$ , so $Q_{enc,2} = Q + (-Q) = 0$ .

$S_3$ encloses all three charges ( $-2Q$ , $Q$ and $-Q$ ), so $Q_{enc,3} = -2Q + Q + (-Q) = -2Q$ .

Step 2 — Apply Gauss's law …
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KCET 2025Set D-41 markMCQ
$Q. Which of the following is a correct statement? (A) Gauss's law is true for any open surface (B) Gauss's law is not applicable when charges are not symmetrically distributed over a closed surface. (C) Gauss's law does not hold good for a charge situated outside the Gaussian surface. (D) Gauss's law is true for any closed surface$
›Reveal solutionSolution

Gauss's law holds for any closed surface regardless of shape or charge symmetry — symmetry is only a computational convenience — so (D) is the correct statement.

Step 1 — State Gauss's law.
$\oint_{S} \vec{E}\cdot d\vec{S} \;=\; \frac{q_{\text{enclosed}}}{\varepsilon_0}$
The surface $S$ must be a closed surface (a Gaussian surface). The law says the net outward flux through it depends only on the total charge enclosed — nothing else.

Step 2 — Test (A): 'true for any open surface'.

The flux integral in Gauss's law is a closed-surface integral ( $\oint$ ). An open surface does not enclose a volume, so 'charge enclosed' is undefined for it. (A) is false.

Step 3 — Test (B): 'not applicable when charges are not symmetrically distributed'.

Gauss's law is a direct consequence of Coulomb's inverse-square law plus the superposition principle; nothing in its derivation assumes symmetry. It is always valid.

Symmetry (spherical, cylindrical, planar) is only what allows us to pull $E$ out of the integral and solve for $\vec{E}$ easily. Without symmetry the law still holds — it is just no longer a convenient tool for finding $\vec{E}$ . Confusing 'not useful for calculation' with 'not applicable' is the trap here. (B) is false.

Step 4 — Test (C): 'does not hold good for a charge situated outside the Gaussian surface'. …
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KCET 2025Set D-41 markMCQ
$Q. Given, a current carrying wire of non-uniform cross-section, which of the following is constant throughout the length of wire? (A) Drift speed (B) Current and drift speed (C) Current only (D) Current, electric field and drift speed$
›Reveal solutionSolution

Conservation of charge forces the current to be the same everywhere; since $I = neAv_d$ with $A$ varying, $v_d$ (and $E$ , and $J$ ) must change from section to section.

Step 1 — Why the current must be constant: charge conservation.

Consider any portion of the wire between two cross-sections. In the steady state, the amount of charge inside that portion does not change with time (charge is not accumulating anywhere — if it were, the resulting electric field would immediately push it out). Therefore:
$\text{charge entering per second} = \text{charge leaving per second}$ $\Longrightarrow \quad I_{\text{in}} = I_{\text{out}}$
So the current $I$ has the same value at every cross-section, no matter how the wire's thickness changes. This is just the equation of continuity, and it is the electrical analogue of "what flows into a pipe must flow out".

Step 2 — The microscopic expression for current.
$I = n\,e\,A\,v_d$
where $n$ = free-electron density (a property of the material, so constant along a wire of one metal), $e$ = electronic charge (a universal constant), $A$ = cross-sectional area, $v_d$ = drift speed.

Step 3 — Deduce how the drift speed behaves.

Rearranging, and using the fact that $I$ , $n$ and $e$ are all constant along the wire:
$v_d = \frac{I}{n e A} \quad\Longrightarrow\quad v_d \propto \frac{1}{A}$
So where the wire is narrow ( $A$ small), the electrons must drift faster; where it is thick, they drift more slowly. Since the cross-section is stated to be non-uniform, $A$ changes, and therefore $v_d$ is not constant.

(Exactly the same intuition as water in a pipe: the flow rate is fixed, so the water speeds up at a constriction.)

Step 4 — What else varies.

The current density is …
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COMEDK 2025Set 2025-A1 markMCQ
$Q. The electric field versus distance graph is shown as given. Select the correct statement from the following. E- electric field R-radius r - distance from the centre (A) The graph shows the variation of the electric field intensity with distance from the centre of a uniformly charged conducting ring of radius R. (B) The graph shows the variation of the electric field intensity with distance from the centre of a uniformly charged non conducting solid sphere of radius R. (C) The graph shows the variation of the electric field intensity with distance from the centre of a uniformly charged conducting solid sphere of radius R. (D) The graph shows the variation of the electric field intensity with distance from the centre of a uniformly charged non conducting cylinder of radius R.$
›Reveal solutionSolution
The graph shows a linear rise inside the object (E ∝ r) up to r = R, then an inverse‑square fall‑off for r > R. This is the signature of a uniformly charged non‑conducting solid sphere, so the correct option is (B).

The key is to match the shape of the graph to the known electric field behaviour of standard charge distributions.

For a conducting solid sphere, all charge resides on the surface, so inside (r < R) the field is zero — the graph would be flat at zero, not rising linearly.

For a conducting ring, the field is zero at the centre, rises to a maximum somewhere off‑centre, then decays — but the rise is not linear from the centre, and the maximum is not at r = R.

For a non‑conducting cylinder, the field inside grows linearly with r only if the cylinder is infinitely long; but the graph shows a clear maximum at r = R and then a decay, which matches a sphere, not a cylinder.

For a uniformly charged non‑conducting solid sphere, Gauss’s law gives exactly:

Inside (r < R): $E = \frac{1}{4\pi\varepsilon_0} \frac{Q r}{R^3}$ → linear in r.

Outside (r > R): $E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$ → inverse‑square decay. The maximum occurs at r = R, where the two expressions match.

Let’s walk through the reasoning step by step.

Identify the key features of the graph

At r = 0, E = 0.

For 0 < r < R, E increases linearly with r (straight line through origin).

At r = R, E reaches a maximum.

For r > R, E decreases smoothly, following a curve that decays like $1/r^2$ (concave up, approaching zero).

Recall the electric field for a uniformly charged non‑conducting solid sphere

By Gauss’s law, for a sphere of radius R with uniform volume charge density ρ:

Inside (r < R): enclosed charge $q_{\text{enc}} = \rho \cdot \frac{4}{3}\pi r^3$ , so

$E \cdot 4\pi r^2 = \frac{q_{\text{enc}}}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{\rho r}{3\varepsilon_0} \propto r.$
- Outside (r > R): total charge $ Q = \rho \cdot \frac{4}{3}\pi R^3 $, so
$E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}.$

The two expressions match at r = R, giving the maximum field there. This exactly reproduces the graph.

Eliminate the other options …
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COMEDK 2025Set 2025-E1 markMCQ
$70 \mu \mathrm{Cm}^{-2}$
›Reveal solutionSolution
The electric flux leaving a closed surface equals the enclosed charge divided by ε₀ (Gauss’s law). For a sphere with given surface charge density, multiply density by surface area to get total charge, then divide by ε₀. The result is about $9.9 \times 10^5\ \mathrm{N\,C^{-1}\,m^2}$ , matching option (A).

Concept & Intuition

Gauss’s law is the star here: the total electric flux through any closed surface is simply $Q_{\text{enc}} / \varepsilon_0$ , independent of the shape. For a conducting sphere, all charge resides on its surface. So the “enclosed charge” is just the total charge on the sphere. We’re given surface charge density $\sigma$ , so we find the sphere’s surface area, multiply to get $Q$ , then apply Gauss’s law. No integration needed — just careful arithmetic.

Step-by-step solution

Find the sphere’s radius and surface area

Diameter = 0.2 m → radius $r = 0.1\ \mathrm{m}$ .

Surface area of a sphere: $A = 4\pi r^2 = 4\pi (0.1)^2 = 4\pi \times 0.01 = 0.04\pi\ \mathrm{m}^2$ .

Numerically, $0.04\pi \approx 0.12566\ \mathrm{m}^2$ .

Compute total charge on the sphere

Surface charge density $\sigma = 70\ \mu\mathrm{C/m^2} = 70 \times 10^{-6}\ \mathrm{C/m^2}$ .

Total charge $Q = \sigma A = (70 \times 10^{-6}) \times (0.04\pi)$ .

$Q = 70 \times 10^{-6} \times 0.04\pi = 2.8 \times 10^{-6} \pi\ \mathrm{C}$ .

Numerically: $2.8 \times 10^{-6} \times 3.1416 \approx 8.796 \times 10^{-6}\ \mathrm{C}$ .

Apply Gauss’s law

Electric flux $\Phi_E = \dfrac{Q}{\varepsilon_0}$ , where $\varepsilon_0 = 8.854 \times 10^{-12}\ \mathrm{C^2/(N\,m^2)}$ . …
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KCET 2024Set D-21 markMCQ
$E = 3 \times 10^5\,\text{NC}^{-1}$
›Reveal solutionSolution

Find the direction of the area vector (normal to the plane), see that it is parallel to $\vec E$ , then use $\phi = \vec E\cdot\vec A = EA$ .

Step 1 — The concept: flux uses the normal, not the plane

Electric flux through a flat area is
$\phi = \vec{E}\cdot\vec{A} = EA\cos\theta$
where $\vec A$ points perpendicular to the surface and $\theta$ is the angle between $\vec E$ and $\vec A$ . The single most common error here is to use the angle between $\vec E$ and the plane instead.

Step 2 — Orient the rectangle

The rectangle's plane is parallel to the Z–X plane. The normal to the Z–X plane is the Y-axis, so
$\vec{A} = A\,\hat{j}$
The field is $\vec{E} = 3\times10^{5}\,\hat{j}\ \text{NC}^{-1}$ — also along $\hat j$ . Hence $\theta = 0^\circ$ and $\cos\theta = 1$ : the field passes straight through the rectangle, giving maximum flux.

Step 3 — Compute the area in SI units …
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KCET 2024Set D-21 markMCQ
$V_0$
›Reveal solutionSolution

Solve each five-resistor half by node/symmetry — each half collapses to a single resistor ( $R$ on the left, $2R$ on the right) — then $I = V_0/(R + 2R) = V_0/3R$ through the single wire joining them.

Step 1 — Write down the left-half network.

Nodes $N_1$ (fed from A) … $N_4$ . The five resistors, all of value $R$ , are:
$N_1\!-\!N_2,\quad N_2\!-\!N_3,\quad N_3\!-\!N_4 \ \ (\text{the main chain}),$ $N_2\!-\!N_4 \ \ (\text{top bridging branch}),\qquad N_1\!-\!N_3 \ \ (\text{bottom bridging branch}).$
Because the two bridging branches overlap (they share the $N_2\!-\!N_3$ resistor), this is not a simple series–parallel chain — it must be solved by node equations (or symmetry). Trying to "parallel" them piecewise is the trap in this question.

Step 2 — Node analysis for the equivalent resistance $N_1 \to N_4$ .

Put $V_1 = V$ at $N_1$ and $V_4 = 0$ at $N_4$ ; let the unknown node potentials be $V_2, V_3$ .

KCL at $N_2$ (its three resistors go to $N_1$ , $N_3$ , $N_4$ ):
$\frac{V_2 - V_1}{R} + \frac{V_2 - V_3}{R} + \frac{V_2 - V_4}{R} = 0 \;\Longrightarrow\; 3V_2 = V_1 + V_3 + V_4 = V + V_3 .$
KCL at $N_3$ (its three resistors go to $N_2$ , $N_1$ , $N_4$ ):
$\frac{V_3 - V_2}{R} + \frac{V_3 - V_1}{R} + \frac{V_3 - V_4}{R} = 0 \;\Longrightarrow\; 3V_3 = V_2 + V + 0 = V + V_2 .$
Step 3 — Solve.

Subtracting the two equations: $3(V_2 - V_3) = V_3 - V_2 \Rightarrow 4(V_2 - V_3) = 0 \Rightarrow V_2 = V_3$ .

Substituting back: $3V_2 = V + V_2 \Rightarrow V_2 = V_3 = V/2$ .

Physical meaning: $V_2 = V_3$ , so no current flows through the middle rung $N_2\!-\!N_3$ — it is a balanced bridge and that resistor can be deleted. (This is forced by the network's symmetry: swapping $N_1\leftrightarrow N_4$ together with $N_2\leftrightarrow N_3$ maps the network onto itself.)

Step 4 — Get the current and hence the half's resistance.

Current leaving $N_1$ goes down two paths, $N_1\!\to\! N_2$ and $N_1\!\to\! N_3$ :

$I_{\text{left}} = \frac{V - V_2}{R} + \frac{V - V_3}{R} = \frac{V/2}{R} + \frac{V/2}{R} = \frac{V}{R}.$ …
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KCET 2023Set A-31 markMCQ
$\vec{E}$
›Reveal solutionSolution

$V_B = V_A - \vec E\cdot\vec d = V_A - Ex$ (path-independent), so the potential energy at B is $U_B = qV_B = q[V_A - Ex]$ .

Step 1 — Relation between field and potential

For a uniform electric field $\vec E$ , the potential difference between two points separated by displacement $\vec d$ is
$V_A - V_B = \vec E \cdot \vec d = E\,x,$
where $x$ is the component of the A→B displacement along $\vec E$ . The electrostatic field is conservative, so the curved path is irrelevant — only the end points matter.

Step 2 — Potential at B
$V_B = V_A - E x .$
This is just the statement "the potential decreases as you move along the direction of $\vec E$ ", by $E$ per unit distance.

Step 3 — Potential energy of the charge at B

By definition, the potential energy of a point charge $q$ at a point of potential $V$ is $U = qV$ . Therefore
$U_B = q\,V_B = q\,[\,V_A - E x\,].$
Step 4 — Check the distractors …
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COMEDK 2023Set 2023-E1 markMCQ
$1 \mathrm{~m}$
›Reveal solutionSolution

New flux = (q/3)/epsilon_0 = (1/3) * (q/epsilon_0) = Phi / 3.

Concept: Gauss's law - the electric flux through a CLOSED surface depends only on the net charge enclosed, not on the size, shape or position of the surface:

Phi = q_enclosed / epsilon_0.

Initially: Phi = q / epsilon_0 (cube of side 1 m). …
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Q.Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region,

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Step 1 — Identify the enclosed charge for each surface

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Step 1 — The concept: flux uses the normal, not the plane

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Step 1 — Relation between field and potential

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