Applications of Gauss's Law

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   By Gauss's law, only the charges enclosed by the surface contribute to the flux; charge $q_3$ outside contributes nothing. $\phi = \dfrac{q_2-q_1}{\varepsilon_0}$

2. 2.↑ back to the question

   According to Gauss's law, $\phi = \dfrac{q_{enclosed}}{\varepsilon_0}$. When the radius of the spherical Gaussian surface is increased, its surface area increases, but the point charge enclosed remains the same. Hence there is no change in the electric flux.

3. 3.↑ back to the question

   By Gauss's law, total charge enclosed by the cube, $q_{en} = 0$ (net charge on a dipole is zero, since $+q$ and $-q$ are both enclosed). $\therefore\ \phi_E = \dfrac{q_{en}}{\varepsilon_0} = 0$

4. 4.↑ back to the question

   (i) Both flat faces have outward flux contributions of the same sign (field diverges outward from the origin on both sides): Net flux, $\phi = \displaystyle\int E\,dA+\int E\,dA = 200\pi(0.05)^2\cos0°\times2 = \pi\,\text{N m}^2\text{C}^{-1}$ (ii) By Gauss's law, net charge enclosed, $q = \phi\,\varepsilon_0 = \pi\,\text{N m}^2\text{C}^{-1}\times8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$ $q = 27.789\times10^{-12}\,\text{C}$

5. 5.↑ back to the question

   (a) Electric flux linked with a surface is the number of electric lines of force cutting through it normally; it is a scalar quantity: $\phi = \displaystyle\int_S\vec{E}\cdot d\vec{S}$. Imagine the given square as one face of a cube of edge $d$. Since the charge $q$ is at distance $d/2$ directly above the centre of the square, it lies at the centre of this imaginary cube, so by Gauss's theorem, the electric flux linked with the whole cube is $\phi = \dfrac{q}{\varepsilon_0}$. By symmetry, this flux is divided equally among the cube's 6 faces, so the flux through the given square is $\phi_{sq} = \dfrac{\phi}{6} = \dfrac{q}{6\varepsilon_0}$. (b) Now the distance of the charge from the plane doubles ($d$) and the side of the square also doubles ($2d$), so the same symmetry holds with a cube of side $2d$ — the charge again sits at the centre of an (enlarged) imaginary cube. Hence the flux through the square is unaffected: $\phi_{sq} = \dfrac{q}{6\varepsilon_0}$, unchanged.

6. 6.↑ back to the question

   Since $\vec{E}=2x\hat{i}$ depends only on $x$, flux passes only through the faces of the cube perpendicular to the $x$-axis. At the left face ($x=0$): $E_L = 0$. At the right face ($x=a$): $E_R = 2a$. Net flux, $\phi = \vec{E}\cdot\Delta\vec{s} = E_L\Delta s\cos180°+E_R\Delta s\cos0° = 0+2a\times a^2 = 2a^3$ Using Gauss's law, $\phi = \dfrac{q}{\varepsilon_0} \ \Rightarrow\ q = \varepsilon_0\phi = 2a^3\varepsilon_0$

7. 7.↑ back to the question

   Electric flux linked with a surface is the number of electric lines of force cutting through the surface normally. Its SI unit is $\text{N m}^2\text{C}^{-1}$ (or V m). Justification: Consider a charge $Q$ placed inside both a sphere and a cube. In each case, by Gauss's theorem the net flux through the closed surface is $\phi_{net} = \dfrac{Q}{\varepsilon_0}$ — the same value regardless of whether the enclosing surface is a sphere, a cube, or any other shape or size, as long as the enclosed charge is the same. This is because the total number of field lines originating from $Q$ is fixed, and they must all cross any closed surface surrounding the charge exactly once.

8. 8.↑ back to the question

   (i) Charge enclosed by $S_1$ is $2Q$, so flux $\phi_1 = \dfrac{2Q}{\varepsilon_0}$. Charge enclosed by $S_2$ is $2Q+4Q=6Q$, so flux $\phi_2 = \dfrac{6Q}{\varepsilon_0}$. Ratio, $\phi_1:\phi_2 = 2:6 = 1:3$ (ii) When a medium of dielectric constant $\varepsilon_r$ is introduced inside $S_1$, the flux through $S_1$ becomes $\phi_1' = \dfrac{2Q}{\varepsilon_0\varepsilon_r}$.

9. 9.↑ back to the question

   Let the charge on the sphere be $q$. Charge induced on the inner surface $=-q$, so $\sigma_{in} = \dfrac{-q}{4\pi r^2}$ Charge on the outer surface $=+q$, so $\sigma_{out} = \dfrac{q}{4\pi(2r)^2} = \dfrac{q}{16\pi r^2}$ $\dfrac{\sigma_{in}}{\sigma_{out}} = \dfrac{-q/4\pi r^2}{q/16\pi r^2} = -4$, i.e., $\sigma_{in}:\sigma_{out} = -4:1$

10. 10.↑ back to the question

    (i) Charge induced on the inner surface $=-Q$, so $\sigma_1 = \dfrac{-Q}{4\pi R_1^2}$ (ii) Charge induced on the outer surface $=+Q$, so $\sigma_2 = \dfrac{Q}{4\pi R_2^2}$

11. 11.↑ back to the question

    No, the charge given to a metallic sphere does not depend on whether it is hollow or solid, because in either case all the excess charge resides on the outer surface of the conductor (the electric field inside a conductor is zero in electrostatic equilibrium).

12. 12.↑ back to the question

    Consider a spherical Gaussian surface of radius $r$ ($r>R$, the shell's radius), concentric with the shell. By symmetry, $\vec{E}$ is radial and has the same magnitude at every point of this surface. Using Gauss's theorem at point $P$ on this surface: $\displaystyle\oint\vec{E}\cdot d\vec{s} = \dfrac{Q}{\varepsilon_0}$ $\displaystyle\int E\,4\pi r^2\cos0° = \dfrac{Q}{\varepsilon_0}$ ($E$ is constant over the surface) $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$ This is exactly the field of a point charge $Q$ placed at the centre — hence the field outside a uniformly charged spherical shell is as if the entire charge were concentrated at its centre.

13. 13.↑ back to the question

    Let the field due to each sheet alone be $E_1=E_2=\dfrac{\sigma}{2\varepsilon_0}$, pointing away from the positive sheet and towards the negative sheet. (i) Between the sheets: both fields point in the same direction (from $+\sigma$ towards $-\sigma$), so they add: $E_{net} = |\vec{E}_1+\vec{E}_2| = \dfrac{\sigma}{2\varepsilon_0}+\dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0}$ (ii) Outside the sheets (on either side): the two fields point in opposite directions and are equal in magnitude, so they cancel: $E_{net} = E_1-E_2 = 0$

14. 14.↑ back to the question

    Using Gauss's theorem, the field at $P_1$ (inside the free-space cavity, at distance $r_1$ from the centre): $E_1 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r_1^2}$ (as if due to the point charge $+Q$ alone). The field at $P_2$ (inside the metal of the shell, at distance $r_2$ from the centre): $E_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r_2^2} \times 0$... rather, since electric field inside a conductor is always zero, $E_2 = 0$.

15. 15.↑ back to the question

    (a) Electric field due to an infinite charged sheet, $E = \dfrac{\sigma}{2\varepsilon_0}$ — this is independent of the distance from the sheet, so it has the same magnitude at every point on either side. All of $P$, $Q$, $R$ (each $50\,\text{cm}$ from the sheet) have the same field magnitude as at $A$, since the field of an infinite sheet does not depend on distance from it. $E = \dfrac{1\times10^{-15}\times4\pi\times9\times10^9}{18\pi\times2}$ (using $\dfrac{1}{\varepsilon_0}=4\pi\times9\times10^9$) $E = 10^{-6}\,\text{N/C}$, directed outward from the sheet. (b) When brought into contact, $q_1'=q_2'=\dfrac{10-20}{2}=-5\,\mu\text{C}$, and new separation $r'=r/2$. $F' = \dfrac{kq_1'q_2'}{r'^2} = \dfrac{k\times5\times5\times10^{-12}}{(r/2)^2} = \dfrac{k\times5\times5\times4\times10^{-12}}{r^2}$ Comparing with $F=\dfrac{k\times10\times20\times10^{-12}}{r^2}$: $\dfrac{F'}{F} = \dfrac{5\times5\times4}{10\times20} = \dfrac{1}{2}$, so $F' = \dfrac{F}{2}$

16. 16.↑ back to the question

    (a) Consider a cylindrical Gaussian surface $S$ of radius $x$ and length $l$, coaxial with the charged wire. The flux through the two flat circular ends is zero (field is perpendicular to their normal), and through the curved surface: $\phi = \displaystyle\oint\vec{E}\cdot d\vec{S} = \int E\,dS\cos90°+\int E\,dS\cos0°+\int E\,dS\cos90° = E\cdot2\pi xl$ By Gauss's theorem, $\phi = \dfrac{q}{\varepsilon_0} = \dfrac{\lambda l}{\varepsilon_0}$ (charge enclosed on length $l$ is $\lambda l$). $\therefore\ E\times2\pi xl = \dfrac{\lambda l}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{\lambda}{2\pi\varepsilon_0 x}$, directed radially outward from the wire. (b) Since $E \propto \dfrac{1}{x}$, the graph of $E$ vs $x$ is a rectangular-hyperbola-like curve, steeply decreasing and approaching zero as $x$ increases.

17. 17.↑ back to the question

    (a) By induction, charge $+q$ is induced on the inner surface (to neutralise the field inside the conductor due to $-q$ at the centre), so the inner surface charge is $+q$: $\sigma_{inner} = \dfrac{q}{4\pi r_1^2}$ Charge conservation: total charge on the conductor is $Q$, so the outer surface carries $(Q-q)$: $\sigma_{outer} = \dfrac{Q-q}{4\pi r_2^2}$ (b) For a point outside the sphere at distance $x>r_2$, using a spherical Gaussian surface, the total charge enclosed is $(Q-q)+q-q = Q-q$... more directly, the net charge enclosed is $(-q)+Q = Q-q$: $\displaystyle\oint\vec{E}\cdot d\vec{S} = \dfrac{Q-q}{\varepsilon_0} \ \Rightarrow\ E\times4\pi x^2 = \dfrac{Q-q}{\varepsilon_0}$ $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q-q}{x^2}$

18. 18.↑ back to the question

    In region II: the field due to sheet $A$ ($\sigma_1$) points from left to right (positive direction), while the field due to sheet $B$ ($\sigma_2$) points from right to left (negative direction). Therefore, in region II: $E = \dfrac{\sigma_1}{2\varepsilon_0}-\dfrac{\sigma_2}{2\varepsilon_0} = \dfrac{1}{2\varepsilon_0}(\sigma_1-\sigma_2)$, along the positive direction (since $\sigma_1>\sigma_2$). In region III: the fields due to both sheets point in the same direction (left to right, positive direction). Therefore, in region III: $E = \dfrac{\sigma_1}{2\varepsilon_0}+\dfrac{\sigma_2}{2\varepsilon_0} = \dfrac{1}{2\varepsilon_0}(\sigma_1+\sigma_2)$, along the positive direction.

19. 19.↑ back to the question

    Gauss's law: the total flux over a closed surface $S$ in vacuum is $\dfrac{1}{\varepsilon_0}$ times the total charge enclosed by that surface: $\phi = \displaystyle\oint_S\vec{E}\cdot d\vec{S} = \dfrac{q_{enclosed}}{\varepsilon_0}$ (Derivation, same as Q57(a).) Using a coaxial cylindrical Gaussian surface of radius $r$ and length $l$: flux through the flat ends is zero, and through the curved surface $\phi = E\times2\pi rl$. By Gauss's theorem, $\phi = \dfrac{\lambda l}{\varepsilon_0}$, so $E\times2\pi rl = \dfrac{\lambda l}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$, directed radially outward (for $\lambda>0$).

20. 20.↑ back to the question

    (i) For a point inside the shell ($x<R$), a spherical Gaussian surface of radius $x$ encloses no charge (all the charge $q=\sigma\times4\pi R^2$ resides on the shell of radius $R>x$). By Gauss's law, $E\times4\pi x^2 = 0 \ \Rightarrow\ E=0$. (ii) For a point outside the shell ($x>R$), take a concentric spherical Gaussian surface of radius $x$. By symmetry $\vec{E}$ is radial and uniform over this surface: $\displaystyle\oint\vec{E}\cdot d\vec{S} = E\times4\pi x^2$ Charge enclosed $= \sigma\times4\pi R^2 = q$ (total charge on shell), so by Gauss's law: $E\times4\pi x^2 = \dfrac{\sigma\times4\pi R^2}{\varepsilon_0} = \dfrac{q}{\varepsilon_0}$ $E = \dfrac{q}{4\pi\varepsilon_0 x^2}$, or vectorially, $\vec{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}\hat{r}$ — identical to the field of the entire charge $q$ concentrated at the centre, confirming the required result. The graph of $E$ vs $r$ is zero for $r<R$, and falls off as $E\propto1/r^2$ for $r>R$ (with a discontinuous jump to $\sigma/\varepsilon_0$ exactly at $r=R$).

21. 21.↑ back to the question

    (a),(b) (Same as Q61.) Inside a charged spherical shell ($r<R$), $E=0$ (no enclosed charge); outside ($r>R$), $E = \dfrac{q}{4\pi\varepsilon_0 r^2}$, falling off as $1/r^2$. At $r=R$, $E=\dfrac{\sigma}{\varepsilon_0}$. (c) Given: side $=10\,\text{cm}$, so $A=0.01\,\text{m}^2$; $E=200\,\text{NC}^{-1}$; angle with the field $=30°$, so angle with the area vector (normal) $=60°$. $\phi = EA\cos\theta = 200\times0.01\times\cos60° = 1.0\,\text{N m}^2/\text{C}$

22. 22.↑ back to the question

    (a),(b) (Same derivation as Q57/60.) $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$, directed radially outward; $E \propto 1/r$, so the graph is a steadily decreasing curve. (c) Work done in moving charge $q$ from $r_2$ to $r_1$ (against/along the field): $W = \displaystyle\int_{r_2}^{r_1}\vec{F}\cdot d\vec{r} = \int_{r_2}^{r_1}\dfrac{q\lambda}{2\pi\varepsilon_0 r}dr\cos0° = \dfrac{q\lambda}{2\pi\varepsilon_0}\int_{r_2}^{r_1}\dfrac{dr}{r}$ $= \dfrac{q\lambda}{2\pi\varepsilon_0}[\ln r]_{r_2}^{r_1} = \dfrac{q\lambda}{2\pi\varepsilon_0}\ln\dfrac{r_1}{r_2}$

23. 23.↑ back to the question

    Assume a cylindrical Gaussian surface $S$ cutting through the plane sheet of charge, such that point $P$ lies on its plane (flat) face. Net electric flux through $S$: $\phi = \displaystyle\oint_S\vec{E}\cdot d\vec{S} = \int_{\text{left face}}E\,dS\cos0°+\int_{\text{curved}}E\,dS\cos90°+\int_{\text{right face}}E\,dS\cos0°$ $\phi = EA+0+EA = 2EA$ But by Gauss's theorem, $\phi = \dfrac{q}{\varepsilon_0} = \dfrac{\sigma A}{\varepsilon_0}$ (where $q$ is the charge in area $A$ of the sheet enclosed by $S$). $\therefore\ 2EA = \dfrac{\sigma A}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{\sigma}{2\varepsilon_0}$ The field is directed normal to the sheet — away from it if positively charged, and towards it if negatively charged.

24. 24.↑ back to the question

    (Same as the $r<R$ case of Q61.) For a point $P$ inside the shell, take a concentric spherical Gaussian surface of radius $r<R$ through $P$. Since all the charge $q$ resides on the shell of radius $R$ (outside this Gaussian surface), the charge enclosed by this surface is zero. By Gauss's law, $E\times4\pi r^2 = \dfrac{0}{\varepsilon_0} = 0 \ \Rightarrow\ E=0$ Hence the electric field at every point inside a uniformly charged spherical shell is zero.

25. 25.↑ back to the question

    Consider a Gaussian spherical surface of radius $x$, centred at $O$. Total flux through this surface: $\phi = \displaystyle\oint E\,ds = E\oint ds = E\times4\pi x^2$ ...(i) (since $\oint ds = 4\pi x^2$) By Gauss's theorem, $\phi = \dfrac{q_{enclosed}}{\varepsilon_0}$ ...(ii) (i) For $r<x<R$: the Gaussian surface encloses only the inner sphere's charge, $q_{enclosed}=q$. From (i) and (ii): $E\times4\pi x^2 = \dfrac{q}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{q}{4\pi\varepsilon_0 x^2}$ (ii) For $x>R$ (outside the shell): the Gaussian surface encloses both charges, $q_{enclosed}=q+Q$. From (i) and (ii): $E\times4\pi x^2 = \dfrac{q+Q}{\varepsilon_0} \ \Rightarrow\ E = \dfrac{q+Q}{4\pi\varepsilon_0 x^2}$

26. 26.Answer: (A)↑ back to the question

    By Gauss's law, $\phi = \dfrac{q_{enclosed}}{\varepsilon_0}$ (or $\dfrac{q_{enclosed}}{\varepsilon}$ in a medium), which depends only on the net charge enclosed and the permittivity of the medium — not on the size or shape of the Gaussian surface.

27. 27.Answer: (C)↑ back to the question

    Field due to point charge $2q$ at distance $r$: $E = \dfrac{K\cdot2q}{r^2}$ ...(i) Field due to shell (behaves as point charge $q$ at its centre, for points outside) at distance $r/2$: $E' = \dfrac{Kq}{(r/2)^2} = \dfrac{Kq\times4}{r^2}$ ...(ii) Dividing (i) by (ii): $\dfrac{E}{E'} = \dfrac{2q}{r^2}\times\dfrac{r^2}{4q} = \dfrac{1}{2}$ $\therefore\ E' = 2E$