Q.Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10^{-22}\,\text{C/m}^2$. What is E:

Showing the 12 most recent of 33 on this concept.

1. 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

   1. By Gauss's law, the field at the surface of a uniformly charged sphere of radius $R$ carrying charge $Q$ is $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$.
   2. Inflating the balloon increases $R$ but does not add or remove any charge, so $Q$ stays constant. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A cylinder of radius 50 cm and length 2m is placed along x-axis as shown in the figure. If electric field is $70\hat i$ along its axis, the electric flux through the left face of the cylinder is (A) 50 (B) 52 (C) 55 (D) 58

   ›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

   1. The left face is a circular disk of radius $r=0.5\ m$, with area $A=\pi r^2$.
   2. Compute the area: $A = \pi\times(0.5)^2 = \pi\times0.25 \approx \dfrac{22}{7}\times0.25 = 0.7857\ m^2$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Two large circular metal plates each of radius 50 cm carrying equal and unlike charges are parallel to each other. If the electric field between the plates is $720\ NC^{-1}$, then the magnitude of charge on any one plate is (A) 7.5 nC (B) 10 nC (C) 2.5 nC (D) 5 nC

   ›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

   1. Field between two oppositely charged plates: $E = \dfrac{\sigma}{\varepsilon_0}$.
   2. Solve for surface charge density: $\sigma = E\varepsilon_0 = 720 \times 8.854\times10^{-12} = 6.3749\times10^{-9}\ C/m^2$.
   3. Area of each circular plate: $A = \pi r^2 = \pi (0.5)^2 = \pi(0.25) \approx 0.7854\ m^2$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Electric flux through cube of side 'a' enclosing charge 'q' is (A) $q/a^2$ (B) $q/\epsilon_0$ (C) Zero (D) $\dfrac{2q}{a\,\epsilon_0}$

   ›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

   1. Gauss's law states $\oint \vec E\cdot d\vec A = \dfrac{q_{enc}}{\epsilon_0}$ for any closed surface.
   2. Here the closed surface is a cube of side $a$ that encloses a charge $q$.
   3. Regardless of the cube's side length $a$ or the position of $q$ inside it, the total flux is simply $\Phi = \dfrac{q}{\epsilon_0}$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The net outward flux through surface of a box is $8.0\times10^3\ \mathrm{Nm^2C^{-1}}$. The net charge inside the box is (approximately) (A) $70\ nC$ (B) $42\ nC$ (C) $21\ nC$ (D) $60\ nC$

   ›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

   1. Given $\Phi=8.0\times10^3\,\mathrm{Nm^2C^{-1}}$.
   2. Rearranging Gauss's law: $q_{enc}=\varepsilon_0\Phi$.
   3. Substitute $\varepsilon_0=8.85\times10^{-12}\,\mathrm{C^2N^{-1}m^{-2}}$: $q_{enc}=8.85\times10^{-12}\times8.0\times10^3=7.08\times10^{-8}\,C$.
   4. Convert: $7.08\times10^{-8}\,C=70.8\,nC\approx70\,nC$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.The electric field intensity on the surface of a charged sphere of radius R and volume charge density ρ is (A) $\dfrac{3\epsilon_0}{R^2}$ (B) $\dfrac{1}{4\pi\epsilon_0}\dfrac{R^2}{\rho}$ (C) $\dfrac{R^2}{3\epsilon_0}$ (D) $\dfrac{\rho R}{3\epsilon_0}$

   ›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

   1. Total charge enclosed by the sphere: $Q=\rho \times \text{Volume} = \rho\cdot\dfrac{4}{3}\pi R^3$.
   2. By Gauss's law, the field at the surface ($r=R$) is $E=\dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{R^2}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Two conducting thin concentric shells of radii r and 2r are shown in figure. Outer shell carries a charge Q. Inner shell is neutral. The charge that will flow from inner shell to earth after closing the switch s is [FIGURE] (two concentric conducting spherical shells, inner of radius r and outer of radius 2r; the outer shell carries charge Q; the inner shell is connected through a switch s to earth/ground) (A) Q (B) $\dfrac{Q}{3}$ (C) $\dfrac{Q}{2}$ (D) $\dfrac{Q}{4}$

   ›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

   1. 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 $Q+q$.
   2. Potential of the inner shell = (its own contribution) + (contribution of outer shell's charges, since inner shell is inside the outer shell):

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

   3. Setting $V=0$ (grounded): $\dfrac{q}{r} = -\dfrac{Q}{2r} \Rightarrow q = -\dfrac{Q}{2}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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

   1. Field just outside any conductor's surface: $E = \dfrac{\sigma}{\varepsilon_0}$ (a standard result from Gauss's law at a conducting boundary).
   2. This expression contains no explicit dependence on the sphere's radius.
   3. Both spheres (radii 5 cm and 10 cm) are given to have the SAME surface charge density $\sigma$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.In a region, the electric field is given by $\bar{E} = (3\hat{i} + 5\hat{j} + 7\hat{k})\ \mathrm{N\,C^{-1}}$. The electric flux through a surface of area $3\ \mathrm{m^2}$ in yz plane is (in SI units) (A) 21 (B) 15 (C) 12 (D) 9

   ›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

   1. Surface lies in the yz-plane $\Rightarrow$ its area vector is $\vec A = 3\hat i\ \mathrm{m^2}$ (magnitude 3, along x).
   2. Given $\vec E = (3\hat i+5\hat j+7\hat k)\ \mathrm{N/C}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
    Q.The electric flux through the surface of a thin spherical shell of radius 6 cm, having a point charge 2 $\mu$C at its center is (A) $36\pi \times 10^5$ N m$^2$ C$^{-1}$ (B) $72\pi \times 10^5$ N m$^2$ C$^{-1}$ (C) $36\pi \times 10^8$ N m$^2$ C$^{-1}$ (D) $72\pi \times 10^3$ N m$^2$ C$^{-1}$

    ›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

    1. Enclosed charge: $q=2\ \mu\text{C} = 2\times10^{-6}$ C.
    2. $\Phi = \dfrac{q}{\varepsilon_0}$. Using $\dfrac1{4\pi\varepsilon_0}=k=9\times10^9$, we get $\dfrac1{\varepsilon_0}=4\pi k$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.A sphere of radius R and charge 'Q' is placed inside an imaginary sphere of radius 2R such that the centres of two spheres coincide. The electric flux linked with the imaginary sphere is (A) $\dfrac{4Q}{\varepsilon_0}$ (B) $\dfrac{2Q}{\varepsilon_0}$ (C) $\dfrac{Q}{\varepsilon_0}$ (D) $\dfrac{Q}{2\varepsilon_0}$

    ›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

    1. The imaginary sphere of radius $2R$ is concentric with, and completely encloses, the charged sphere of radius $R$ and charge $Q$.
    2. Total charge enclosed by the imaginary sphere: $Q_{enc} = Q$ (no additional charge outside the inner sphere but inside the imaginary sphere). …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
    Q.As shown in the figure, a surface encloses an electric dipole with charge $\pm 6 \times 10^{-6}$ C. The total electric flux through the closed surface is [FIGURE] (an oval closed surface enclosing an electric dipole shown as a rod with $+Q$ labelled at its left end and $-Q$ labelled at its right end) (A) $+12 \times 10^{-6}\ \text{Nm}^2\text{C}^{-1}$ (B) $-12 \times 10^{-6}\ \text{Nm}^2\text{C}^{-1}$ (C) Zero (D) $+6 \times 10^{-6}\ \text{Nm}^2\text{C}^{-1}$

    ›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.

    1. 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.

    2. 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.

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

    3. 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. …

    🔒 Start your 14-day free trial to see the full solution →