Q.Four different closed surfaces, of different shapes and different sizes, are considered. Each one of the four surfaces encloses one and the same single point charge $+q$ (and no other charge). Consider the electric flux through each surface.

Showing the 12 most recent of 16 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.A charge of $5 \mu \mathrm{C}$ is placed at the centre of a spherical shell $S_1$ of radius 10 cm . Now this system is enclosed inside another spherical shell $S_2$ of radius 20 cm . The ratio of the electrical flux through the surface $S_2$ to $S_1$ is : (A) $1: 2$ (B) $4:1$ (C) $2:1$ (D) $1: 1$

   ›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

   1. 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 $S_1$.

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

   2. Flux through $S_1$ By Gauss’s law:

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

   3. 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 $5\,\mu\text{C}$.

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

   4. Flux through $S_2$ Again by Gauss’s law:

   $\Phi_2 = \frac{Q_{\text{enc},2}}{\varepsilon_0} = \frac{5\times10^{-6}}{\varepsilon_0}$

   5. Ratio of fluxes

   $\frac{\Phi_2}{\Phi_1} = \frac{5\times10^{-6}/\varepsilon_0}{5\times10^{-6}/\varepsilon_0} = 1$

   So the ratio is $1:1$.

   Watch out

   A common mistake is to think flux depends on the radius of the shell. It does not — only the enclosed charge matters. The electric field strength changes with radius, but the total flux (field × area, integrated) stays constant.

   Tip

   If you ever forget: imagine a point charge. Draw a small sphere and a large sphere around it. The field is weaker on the larger sphere, but the area is larger — the product (flux) is the same. That’s the geometric essence of Gauss’s law.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. COMEDK 2026Set 2026-A1 markMCQ
   Q.A pith ball of mass ' $m$ ' gram and charge ' $Q$ ' is suspended using a mass less silk thread near a large charged conducting metal sheet of area ' $A$ ' and surface charged density ' $\sigma$ '. If the silk thread makes an angle $\Theta$ with the metal sheet, then: (A) $\tan \theta \propto \sigma$ (B) $\tan \theta \propto A$ (C) $\tan \theta \propto m g$ (D) $\tan \theta \propto \frac{1}{\sigma}$

   ›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

   $\tan\theta=\frac{F}{mg}=\frac{Q\sigma}{\varepsilon_0\, mg}.$

   With $Q$, $m$ and $g$ fixed, the right-hand side is directly proportional to $\sigma$ and does not depend on the sheet area $A$. Hence

   $\tan\theta\propto\sigma.$

   ✓Final answer

   $\tan\theta\propto\sigma$ — option (A).

3. COMEDK 2026Set 2026-M1 markMCQ
   Q.A uniform electric field $E=3 \hat{i}+6 \hat{j}+\hat{k}$ passes through a closed cuboidal surface. One face of the cuboid has an area $4 m^2$ and an outward unit normal given by $\frac{2 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{17}}$. If the electric flux through the remaining 5 faces is zero, the charge enclosed by the cuboid is: (A) Cannot be determined (B) $\frac{84 \epsilon_0}{\sqrt{17}}$ (C) zero (D) $\frac{\sqrt{17}}{84 \epsilon_0}$

   ›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

   1. 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

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

   2. 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

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

   3. Apply Gauss’s law The total flux through the closed cuboidal surface is

   $\Phi_{\text{total}} = \frac{Q_{\text{enc}}}{\varepsilon_0}.$

   Since flux through the other five faces is zero,

   $\Phi_{\text{total}} = \Phi = \frac{84}{\sqrt{17}}.$

   Therefore,

   $\frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{84}{\sqrt{17}} \quad\Rightarrow\quad Q_{\text{enc}} = \frac{84\,\varepsilon_0}{\sqrt{17}}.$

   Watch out

   A common mistake is to forget that the given normal is outward — but here it’s already outward, so no sign change. Also, don’t confuse the unit normal with the area vector; you must multiply by the area.

   Tip

   The dot product $\vec{E}\cdot\vec{A}$ directly gives flux in SI units (N·m²/C). No need to integrate because the field is uniform and the face is flat.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. KCET 2026Set C21 markMCQ
   Q.Consider three point charges $-2Q, Q$ and $-Q$ and three surfaces $S_1$, $S_2$ and $S_3$ as shown in the figure . Match the entries of List-I with that of List-II. List-I

   (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

   By Gauss's law, $\Phi = \dfrac{Q_{enc}}{\epsilon_0}$, and any charge outside a given surface contributes nothing to its flux. Therefore:

   $\Phi_1 = \frac{-Q}{\epsilon_0}, \qquad \Phi_2 = 0, \qquad \Phi_3 = \frac{-2Q}{\epsilon_0}$

   So (a) matches (ii), (b) matches (iii), and (c) matches (i).

   ✓Final answer

   The correct option is (D) — a - ii, b - iii, c - i.

5. 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'.

   A charge outside the surface contributes zero net flux: its field lines enter the closed surface on one side and leave on the other, so the inward and outward flux cancel exactly. The law still holds perfectly well — the external charge simply does not appear in $q_{\text{enc}}$ (although it does contribute to $\vec{E}$ at individual points on the surface). (C) is false.

   Step 5 — Test (D): 'true for any closed surface'.

   By the argument above, for any closed surface — sphere, cube, or an arbitrary irregular blob — the net flux equals $q_{\text{enc}}/\varepsilon_0$. This is precisely the content of the law. (D) is TRUE.

   ✓Final answer

   The correct option is (D) — Gauss's law is true for any closed surface.

   ANSWER: D

6. 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

   $J = \frac{I}{A} \propto \frac{1}{A} \qquad \text{— not constant}$

   and from the microscopic Ohm's law $\vec{J} = \sigma\vec{E}$,

   $E = \frac{J}{\sigma} = \frac{I}{\sigma A} \propto \frac{1}{A} \qquad \text{— also not constant}$

   So the electric field inside the wire is stronger in the narrow parts — which is precisely what is needed to accelerate the electrons up to the higher drift speed there.

   Step 5 — Evaluate the options.

   • (A) Drift speed — varies ($\propto 1/A$). ✗
   • (B) Current and drift speed — current is constant, but drift speed is not. ✗
   • (C) Current only — ✓ the only quantity in the list that is genuinely the same throughout.
   • (D) Current, electric field and drift speed — both $E$ and $v_d$ vary. ✗

   ✓Final answer

   The correct option is (C) — Current only.

   ANSWER: C

7. 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.

   1. 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).

   2. 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.

   3. Eliminate the other options

      • (A) Conducting ring: The field along the axis of a ring is zero at the centre, rises to a maximum at $r = R/\sqrt{2}$ (not at R), and then decays. The rise is not linear from the centre.
      • (C) Conducting solid sphere: All charge is on the surface, so inside (r < R) E = 0. The graph would be zero up to r = R, then jump to a maximum and decay — no linear rise.
      • (D) Non‑conducting cylinder: For an infinitely long uniformly charged cylinder, inside E ∝ r, but outside E ∝ 1/r (not 1/r²). The graph shows a 1/r² tail, so it cannot be a cylinder. For a finite cylinder, the field is more complicated and does not have a simple linear‑then‑inverse‑square form.

   4. Conclusion

      Only the uniformly charged non‑conducting solid sphere gives a linear increase from zero at the centre to a maximum at the surface, followed by an inverse‑square decay.

   Watch out

   A common mistake is to confuse the conducting sphere (field zero inside) with the non‑conducting sphere (field linear inside). The graph’s linear rise from the origin rules out any conductor.

   Tip

   Remember the mnemonic: “Inside a uniform sphere, field grows like r; outside, it falls like 1/r².” The peak always occurs at the surface r = R.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. COMEDK 2025Set 2025-E1 markMCQ
   Q.A uniformly charged conducting sphere of 0.2 m diameter has a surface charge density of $70 \mu \mathrm{Cm}^{-2}$. The electric flux leaving the surface of the sphere is: (A) $9.9 \times 10^5 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$ (B) $9.9 \times 10^6 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$ (C) $8.9 \times 10^5 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$ (D) $8.9 \times 10^6 \mathrm{~N} \mathrm{C}^{-1} \mathrm{~m}^2$

   ›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

   1. 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, $0.04\pi \approx 0.12566\ \mathrm{m}^2$.

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

   3. Apply Gauss’s law

      Electric flux $\Phi_E = \dfrac{Q}{\varepsilon_0}$, where $\varepsilon_0 = 8.854 \times 10^{-12}\ \mathrm{C^2/(N\,m^2)}$.

      $\Phi_E = \dfrac{8.796 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 9.93 \times 10^5\ \mathrm{N\,C^{-1}\,m^2}$.

   4. Match to the options

      $9.93 \times 10^5$ rounds to $9.9 \times 10^5$, which is option (A).

   Tip

   A common shortcut: flux = $\sigma A / \varepsilon_0$. Since $A = 4\pi r^2$, the $\pi$ cancels nicely if you keep it symbolic: $\Phi = \frac{70\times10^{-6} \times 4\pi (0.1)^2}{\varepsilon_0} = \frac{70\times10^{-6} \times 0.04\pi}{8.854\times10^{-12}}$. The $\pi$ doesn’t cancel here, but the arithmetic is still clean.

   Watch out

   A classic mistake is to use the sphere’s volume instead of surface area, or to forget that $\mu$ means $10^{-6}$. Double-check units: surface charge density is per area, not per volume.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

9. KCET 2024Set D-21 markMCQ
   Q.A uniform electric field $E = 3 \times 10^5\,\text{NC}^{-1}$ is acting along the positive Y-axis. The electric flux through a rectangle of area $10\,\text{cm} \times 30\,\text{cm}$ whose plane is parallel to the Z-X plane is (A) $12 \times 10^3\,\text{Vm}$ (B) $9 \times 10^3\,\text{Vm}$ (C) $15 \times 10^3\,\text{Vm}$ (D) $18 \times 10^3\,\text{Vm}$

   ›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

   $A = (10\ \text{cm})\times(30\ \text{cm}) = (0.10\ \text{m})\times(0.30\ \text{m}) = 3\times10^{-2}\ \text{m}^2$

   Step 4 — Compute the flux

   $\phi = EA = \left(3\times10^{5}\ \text{NC}^{-1}\right)\left(3\times10^{-2}\ \text{m}^2\right)$

   $\phi = 9\times10^{3}\ \text{N m}^2\text{C}^{-1} = 9\times10^{3}\ \text{V m}$

   (Recall $1\ \text{N m}^2\text{C}^{-1} \equiv 1\ \text{V m}$.)

   ✓Final answer

   The correct option is (B) — $9 \times 10^3\,\text{Vm}$.

   ANSWER: B

10. KCET 2024Set D-21 markMCQ
    Q.In the circuit shown, the end A is at potential $V_0$ and end B is grounded. The electric current I indicated in the circuit is (A) $\dfrac{V_0}{R}$ (B) $\dfrac{2V_0}{R}$ (C) $\dfrac{3V_0}{R}$ (D) $\dfrac{V_0}{3R}$

    ›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 $N_1\!\to\! N_3$:

    $I_{\text{left}} = \frac{V - V_2}{R} + \frac{V - V_3}{R} = \frac{V/2}{R} + \frac{V/2}{R} = \frac{V}{R}.$

    $\therefore\ R_{\text{left}} = \frac{V}{I_{\text{left}}} = \boxed{R}$

    (Equivalently, with the middle rung dead the network is two parallel paths $N_1\!-\!N_2\!-\!N_4$ and $N_1\!-\!N_3\!-\!N_4$, each of $2R$: $2R \parallel 2R = R$. ✓)

    Step 5 — The right half.

    It has identical topology with every resistor equal to $2R$. Resistance scales linearly with a common factor, so

    $R_{\text{right}} = 2R .$

    Step 6 — Combine and find $I$.

    The two halves are joined by the single wire on which $I$ is marked, so they are in series between A ($+V_0$) and B (ground):

    $R_{AB} = R_{\text{left}} + R_{\text{right}} = R + 2R = 3R .$

    That single connecting wire carries the entire current, so

    $I = \frac{V_0 - 0}{R_{AB}} = \frac{V_0}{3R}.$

    ✓Final answer

    The correct option is (D) $\dfrac{V_0}{3R}$ — each balanced-bridge half collapses to $R$ and $2R$ respectively, giving $R_{AB} = 3R$.

    ANSWER: D

11. KCET 2023Set A-31 markMCQ
    Q.A uniform electric field vector $\vec{E}$ exists along horizontal direction as shown. The electric potential at A is $V_A$. A small point charge $q$ is slowly taken from A to B along the curved path as shown. The potential energy of the charge when it is at point B is (A) $q[V_A + E x]$ (B) $q[E x - V_A]$ (C) $q E x$ (D) $q[V_A - E x]$

    ›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

    • (C) $qEx$ ignores the potential $V_A$ at the starting point entirely.
    • (A) $q[V_A + Ex]$ has the potential increasing along $\vec E$ — wrong sign.
    • (B) $q[Ex - V_A]$ is the negative of the correct expression.

    (Note that the charge is moved slowly, so its kinetic energy stays zero and all the work done goes into potential energy — but the answer $U_B = qV_B$ follows from the definition regardless.)

    ✓Final answer

    The correct option is (D) — $q[V_A - E x]$.

    ANSWER: D

12. COMEDK 2023Set 2023-E1 markMCQ
    Q.The electric flux from cube of side $1 \mathrm{~m}$ is '$\Phi$' When the side of the cube is made $3 \mathrm{~m}$ and the charge enclosed by the cube is made one third of the original value, then the flux from the bigger cube will be : (A) $\frac{\Phi}{3}$ (B) $\Phi$ (C) $3 \Phi$ (D) $9 \Phi$

    ›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).

    Now the cube's side becomes 3 m - this changes nothing, since flux is independent of the size of the Gaussian surface - and the enclosed charge becomes q/3.

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

    ✓Final answer

    The correct option is (A) — $\frac{\Phi}{3}$

    ANSWER: A