NCERT Exemplar Solutions Class 12 Physics Electric Charges and Fields — Very Short, Short and Long Answer Questions Q14–31

NCERT Exemplar Problems, Class 12 Physics — Electric Charges and Fields, Very Short Answer Questions 14 to 19, Short Answer Questions 20 to 25 and Long Answer Questions 26 to 31. All 18 questions solved. The book numbers them 1.14 to 1.31.

Almost everything on this page rests on four ideas: Coulomb’s law with the superposition of forces, Gauss’s law $\oint \vec E\cdot d\vec A = q_{\text{enc}}/\varepsilon_0$, symmetry (which tells you which components of a field cancel before you calculate anything), and the binomial approximation for small displacements. The long answer questions mostly combine two of these.

Key insight. Before calculating a field, ask what the symmetry already forces. In Questions 25, 30 and 31 the sideways or in-plane components cancel at once, and a restoring force proportional to the displacement ($F = -kx$) is all you need to claim simple harmonic motion. The period then follows from $T = 2\pi\sqrt{m/k}$.

Very Short Answer Questions

Question 14

An arbitrary surface encloses a dipole. What is the electric flux through this surface?

Solution. By Gauss’s law the flux depends only on the net charge enclosed. A dipole holds $+q$ and $-q$, so the net charge is zero. The flux is therefore zero, whatever the shape of the surface. (The field itself is not zero; as much flux leaves the surface near the $+q$ as enters near the $-q$.)

Question 15

A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. A charge $Q$ is placed at the centre of the spherical cavity. What will be the surface charge density on (i) the inner surface, and (ii) the outer surface?

Solution. The field inside the metal must be zero. Take a Gaussian sphere inside the metal. It must enclose zero net charge, so the inner surface carries an induced charge $-Q$ to cancel the $+Q$ at the centre. The shell itself is neutral, so the charge $+Q$ is pushed to the outer surface. Each charge spreads uniformly because of the spherical symmetry.

$$\sigma_{\text{inner}} = \frac{-Q}{4\pi R_1^2}, \qquad \sigma_{\text{outer}} = \frac{+Q}{4\pi R_2^2}$$

Question 16

The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then, is the electrostatic field inside a conductor zero?

Solution. The strong fields inside atoms are real, but they bind each atom into an electrically neutral unit. When we say the field inside a conductor is zero we mean the macroscopic average over a region containing very many atoms, not the field at a point inside one atom. In an isolated conductor in equilibrium there is no excess charge inside, and free charges rearrange until the average field is zero; otherwise they would keep moving.

Question 17

If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero?

Solution. Zero net charge means zero net flux, not zero field. Flux entering over one part of the surface can balance flux leaving over another; a dipole enclosed by a sphere is the standard example, with a non-zero field at every point of the sphere. So the first statement does not follow.

Conversely, if $\vec E = 0$ at every point of the surface, the flux through it is zero, so $q_{\text{enc}} = \varepsilon_0 \times 0 = 0$. The second statement does follow.

Question 18

Sketch the electric field lines for a uniformly charged hollow cylinder shown in Fig 1.8.

Solution. The figure is a hollow cylinder, positively charged over its curved surface and both end caps. Describe the lines in two views.

• Top view (looking down the axis): the lines run radially outward from the curved surface, like the spokes of a wheel, never crossing.
• Side view: the lines leave the curved surface normally, that is perpendicular to the wall, and run straight outward. Near the two ends they fan out, spreading as they leave the caps, so they are further apart there. Far away the pattern looks like that of a point charge.
• Inside the hollow there is no charge, so there are no field lines in the interior.

Question 19

What will be the total flux through the faces of the cube (Fig. 1.9) with side of length $a$ if a charge $q$ is placed at (a) A: a corner of the cube, (b) B: mid-point of an edge of the cube, (c) C: centre of a face of the cube, (d) D: mid-point of B and C?

Solution. The method is to complete the symmetry by fitting identical cubes around the charge until it sits inside a closed surface. The total flux through all of them is $q/\varepsilon_0$, and each cube takes an equal share.

• (a) Corner: eight cubes meet at a corner, so the cube gets $\dfrac{1}{8}\cdot\dfrac{q}{\varepsilon_0}$.
• (b) Mid-point of an edge: four cubes share an edge, so the cube gets $\dfrac{1}{4}\cdot\dfrac{q}{\varepsilon_0}$.
• (c) Centre of a face: two cubes share a face, so the cube gets $\dfrac{1}{2}\cdot\dfrac{q}{\varepsilon_0}$.
• (d) B is on an edge of the right-hand face and C is the centre of that same face, so their mid-point D also lies on that face. The situation is the same as (c).

Short Answer Questions

Question 20

A paisa coin is made up of Al-Mg alloy and weighs $0.75$ g. It has a square shape and its diagonal measures $17$ mm. It is electrically neutral and contains equal amounts of positive and negative charges. Treating the paisa coins made up of only Al, find the magnitude of equal number of positive and negative charges. What conclusion do you draw from this magnitude?

Solution. The shape and the diagonal are not needed; only the mass matters. Aluminium has molar mass $26.9815$ g and atomic number $13$, so each atom has $13$ protons and $13$ electrons.

Number of atoms:

$$N = \frac{0.75}{26.9815} \times 6.02\times 10^{23} \approx 1.67 \times 10^{22}$$

Charge of each sign:

$$q = 13\,N\,e = 13 \times 1.67\times 10^{22} \times 1.6\times 10^{-19}\ \text{C} \approx 3.48 \times 10^{4}\ \text{C}$$

Conclusion. A tiny coin holds about $34.8$ kC of positive charge and the same amount of negative charge. Ordinary neutral matter therefore contains a huge amount of charge of each sign, which merely cancels exactly.

Question 21

Consider a coin of Example 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8$ kC. Suppose that these equal charges were concentrated in two point charges separated by (i) $1$ cm ($\sim \tfrac12 \times$ diagonal of the one paisa coin), (ii) $100$ m ($\sim$ length of a long building), and (iii) $10^6$ m (radius of the earth). Find the force on each such point charge in each of the three cases. What do you conclude from these results?

Solution. Use Coulomb’s law with $q = 3.48\times 10^{4}$ C, so $kq^2 = 9\times 10^{9} \times (3.48\times10^4)^2 \approx 1.1\times 10^{19}$ N m$^2$.

$$F = \frac{kq^2}{r^2}$$

• (i) $r = 10^{-2}$ m: $F = \dfrac{1.1\times10^{19}}{10^{-4}} \approx 1.1\times 10^{23}$ N
• (ii) $r = 100$ m: $F = \dfrac{1.1\times10^{19}}{10^{4}} \approx 1.1\times 10^{15}$ N
• (iii) $r = 10^{6}$ m: $F = \dfrac{1.1\times10^{19}}{10^{12}} \approx 1.1\times 10^{7}$ N

Conclusion. Even at the radius of the earth the force is of the order of ten million newtons. Forces this large mean that separating even a small fraction of the positive and negative charge in matter is extremely hard: electrical neutrality is very difficult to disturb, and bulk matter is neutral to a very high degree.

Question 22

Fig. 1.10 represents a crystal unit of caesium chloride, CsCl. The caesium atoms, represented by open circles, are situated at the corners of a cube of side $0.40$ nm, whereas a Cl atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron. (i) What is the net electric field on the Cl atom due to eight Cs atoms? (ii) Suppose that the Cs atom at the corner A is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?

Solution.

(i) The eight Cs$^+$ ions sit at the corners of a cube, all at the same distance from the centre. They pair up in diametrically opposite pairs, and in each pair the two fields at the centre are equal and opposite. By symmetry the net field is zero.

(ii) With all eight present the net force on Cl$^-$ is zero. Removing the Cs$^+$ at A therefore leaves a net force equal and opposite to the force that the ion at A would have exerted. That is the same as adding a charge $-e$ at A. The ion at A would have attracted Cl$^-$ towards A, so the force now is directed away from A.

The distance from a corner to the centre is half the body diagonal:

$$r = \frac{\sqrt3}{2}(0.40\ \text{nm}) = 0.346\ \text{nm}$$

$$F = \frac{e^2}{4\pi\varepsilon_0 r^2} = \frac{9\times10^{9}\,(1.6\times10^{-19})^2}{(0.346\times10^{-9})^2} \approx 1.92\times 10^{-9}\ \text{N}$$

Question 23

Two charges $q$ and $-3q$ are placed fixed on the $x$-axis separated by distance $d$. Where should a third charge $2q$ be placed such that it will not experience any force?

Solution. The third charge feels zero force only where the fields of the two charges cancel. Between the charges the fields point the same way, so there is no solution there. Outside, the field of the smaller charge $q$ must balance that of the larger charge $3q$, which is possible only on the side of the smaller charge, away from $-3q$. Let the point be a distance $x$ from $q$, so it is $d + x$ from $-3q$:

$$\frac{q}{x^2} = \frac{3q}{(d+x)^2} \;\Rightarrow\; (d+x)^2 = 3x^2 \;\Rightarrow\; d + x = \sqrt3\,x$$

$$x = \frac{d}{\sqrt3 – 1} = \frac{(\sqrt3+1)\,d}{2}$$

The size and sign of the third charge do not matter, since it cancels from the condition.

Question 24

Fig. 1.11 shows the electric field lines around three point charges A, B and C. (a) Which charges are positive? (b) Which charge has the largest magnitude? Why? (c) In which region or regions of the picture could the electric field be zero? Justify your answer. (i) near A, (ii) near B, (iii) near C, (iv) nowhere.

Solution. The figure shows lines leaving A and C and lines ending on B, with the densest bundle running between B and C.

(a) Field lines start on positive charges and end on negative ones. Lines leave A and C, so A and C are positive; B is negative.

(b) The number of field lines drawn at a charge is proportional to its magnitude. C has the most lines on it, so it has the largest magnitude.

(c) A point where the field is zero (a neutral point) can exist only where the fields of two charges point in opposite directions with equal size. For two like charges this happens on the line joining them, closer to the smaller one. A and C are both positive, and A is the smaller, so the field can vanish near A. Near B, and near C, the nearby charge dominates.

Question 25

Five charges, $q$ each, are placed at the corners of a regular pentagon of side $a$ (Fig. 1.12). (a)(i) What will be the electric field at O, the centre of the pentagon? (ii) What will be the electric field at O if the charge from one of the corners (say A) is removed? (iii) What will be the electric field at O if the charge $q$ at A is replaced by $-q$? (b) How would your answer to (a) be affected if the pentagon is replaced by an $n$-sided regular polygon with charge $q$ at each of its corners?

Solution. Each corner is at the same distance $r$ from O, so each charge produces a field of the same magnitude $\dfrac{q}{4\pi\varepsilon_0 r^2}$, and the five vectors are spaced $72^\circ$ apart.

(a)(i) Five equal vectors spread evenly around a circle sum to zero.

(ii) The total with all five present is zero, so removing A leaves the negative of A’s field. The field of A at O points away from A, so the remaining field points towards A, along OA, with magnitude $\dfrac{q}{4\pi\varepsilon_0 r^2}$.

(iii) Replacing $q$ by $-q$ changes the field at O from $+\vec E_A$ to $-\vec E_A$, a change of $-2\vec E_A$. The net field is therefore $-2\vec E_A$: magnitude $\dfrac{2q}{4\pi\varepsilon_0 r^2}$, again along OA (towards A).

(b) Nothing changes in kind. For a regular $n$-gon, $n$ equal vectors evenly spaced in angle also sum to zero, so every result in (a) holds for any regular polygon, with the same expressions in terms of $r$.

Long Answer Questions

Question 26

In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density $N$, which is maintained a constant. Let the charge on the proton be $e_p = -(1+y)e$ where $e$ is the electronic charge. (a) Find the critical value of $y$ such that expansion may start. (b) Show that the velocity of expansion is proportional to the distance from the centre.

Solution.

(a) Each hydrogen atom has one electron (charge $e$) and one proton ($-(1+y)e$), so its net charge has magnitude $ye$. Treat the Universe as a uniform sphere of atoms with $N$ per unit volume. Take an atom at distance $R$ from the centre. By Gauss’s law only the charge inside radius $R$ pulls on it:

$$E\,(4\pi R^2) = \frac{1}{\varepsilon_0}\cdot\frac{4}{3}\pi R^3\,N\,ye \;\Rightarrow\; E = \frac{N\,y\,e\,R}{3\varepsilon_0}$$

The outward electric force on the atom is $F_e = ye\,E = \dfrac{N y^2 e^2 R}{3\varepsilon_0}$. The inward gravitational pull of the same enclosed matter, of mass $\tfrac43\pi R^3 N m_p$ (the electron mass is negligible), gives $g = \dfrac{4}{3}\pi G N m_p R$ and a force $F_g = m_p g = \dfrac{4\pi}{3}\,G N m_p^2 R$. Expansion can start when the repulsion first equals the attraction:

$$\frac{N y^2 e^2}{3\varepsilon_0} = \frac{4\pi}{3}\,G N m_p^2 \;\Rightarrow\; y_c = \frac{m_p\sqrt{4\pi\varepsilon_0 G}}{e}$$

With $m_p = 1.67\times10^{-27}$ kg, $\sqrt{4\pi\varepsilon_0 G} \approx 8.6\times10^{-11}$ and $e = 1.6\times10^{-19}$ C this gives $y_c \approx 9\times10^{-19} \sim 10^{-18}$. A proton only needs to differ from the electron’s charge by about one part in $10^{18}$.

(b) For $y$ above $y_c$ the net outward force per unit mass is proportional to $R$, so the equation of motion has the form

$$\frac{d^2R}{dt^2} = \alpha^2 R, \qquad \alpha^2 = \frac{N}{3 m_p}\left(\frac{y^2e^2}{\varepsilon_0} – 4\pi G m_p^2\right)$$

whose expanding solution is $R = A e^{\alpha t}$. Differentiating, $\dfrac{dR}{dt} = \alpha R$, so the speed of an atom is proportional to its distance from the centre. This is a Hubble-type law.

Question 27

Consider a sphere of radius $R$ with charge density distributed as $\rho(r) = kr$ for $r \le R$, and $\rho = 0$ for $r > R$. (a) Find the electric field at all points $r$. (b) Suppose the total charge on the sphere is $2e$ where $e$ is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.

Solution.

(a) The charge in a thin shell of radius $r’$ and thickness $dr’$ is $kr’\cdot 4\pi r’^2\,dr’$. The charge inside radius $r \le R$ is

$$Q(r) = 4\pi k\int_0^r r’^3\,dr’ = \pi k\,r^4$$

Gauss’s law over a sphere of radius $r$ gives $E\,4\pi r^2 = Q/\varepsilon_0$.

• For $r \le R$: $\;E = \dfrac{\pi k r^4}{4\pi\varepsilon_0 r^2} = \dfrac{k r^2}{4\varepsilon_0}$, radially outward for $k>0$.
• For $r > R$: all the charge $\pi k R^4$ is enclosed, so $E = \dfrac{kR^4}{4\varepsilon_0 r^2}$.

(b) The total charge is $\pi k R^4 = 2e$, so $k = \dfrac{2e}{\pi R^4}$ (negative, since $e<0$). Inside, $E = \dfrac{e\,r^2}{2\pi\varepsilon_0 R^4}$, and it points towards the centre. Place the two protons on a diameter, a distance $r$ on each side of the centre. On each proton the distribution attracts it towards the centre, with force $|e|E$, and the other proton repels it with force $\dfrac{e^2}{4\pi\varepsilon_0(2r)^2}$. Setting them equal:

$$\frac{e^2 r^2}{2\pi\varepsilon_0 R^4} = \frac{e^2}{16\pi\varepsilon_0 r^2} \;\Rightarrow\; r^4 = \frac{R^4}{8} \;\Rightarrow\; r = \frac{R}{8^{1/4}} \approx 0.59\,R$$

Question 28

Two fixed, identical conducting plates ($\alpha$ and $\beta$), each of surface area $S$ are charged to $-Q$ and $q$, respectively, where $Q > q > 0$. A third identical plate ($\gamma$), free to move is located on the other side of the plate with charge $q$ at a distance $d$ (Fig 1.13). The third plate is released and collides with the plate $\beta$. Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst $\beta$ and $\gamma$. (a) Find the electric field acting on the plate $\gamma$ before collision. (b) Find the charges on $\beta$ and $\gamma$ after the collision. (c) Find the velocity of the plate $\gamma$ after the collision and at a distance $d$ from the plate $\beta$.

Solution. The figure shows three parallel plates in the order $\alpha, \beta, \gamma$, carrying $-Q$, $q$ and $Q$ respectively, with $\gamma$ at distance $d$ from $\beta$ on the right. Each large plate produces a uniform field $\sigma/2\varepsilon_0 = \text{charge}/2\varepsilon_0 S$ on either side, and the field acting on a plate comes from the other plates, not its own charge.

(a) At $\gamma$: the plate $\alpha$ ($-Q$) attracts it to the left, giving $\dfrac{Q}{2\varepsilon_0 S}$; the plate $\beta$ ($+q$) repels it to the right, giving $\dfrac{q}{2\varepsilon_0 S}$.

$$E = \frac{Q – q}{2\varepsilon_0 S}\ \text{directed towards } \beta\ (\text{the } -x \text{ direction})$$

(b) On contact, $\beta$ and $\gamma$ form one conductor. Inside the thickness of the two touching plates the field is zero, the field of $\alpha$ ($Q/2\varepsilon_0 S$) must be cancelled by the difference of the fields of the two plates: $q_1 – q_2 = Q$, where $q_1$ and $q_2$ are the final charges on $\beta$ and $\gamma$. Charge conservation gives $q_1 + q_2 = Q + q$. Hence

$$q_1 = Q + \frac{q}{2}\ (\text{on } \beta), \qquad q_2 = \frac{q}{2}\ (\text{on } \gamma)$$

(c) Work done before the collision, over distance $d$ at constant force $Q E$:

$$W_1 = Q\cdot\frac{(Q-q)}{2\varepsilon_0 S}\,d$$

After the elastic collision $\gamma$ carries $q/2$ and is repelled from $\beta$ (it is pushed in the $+x$ direction). The field on it is $\dfrac{Q+q/2}{2\varepsilon_0 S}$ (repulsion from $\beta$) minus $\dfrac{Q}{2\varepsilon_0 S}$ (attraction to $\alpha$) $= \dfrac{q}{4\varepsilon_0 S}$ along $+x$, so the force is $\dfrac{q}{2}\cdot\dfrac{q}{4\varepsilon_0 S} = \dfrac{q^2}{8\varepsilon_0 S}$ and the work over $d$ is $W_2 = \dfrac{q^2 d}{8\varepsilon_0 S}$. The total, since the elastic collision loses no kinetic energy:

$$\tfrac12 m v^2 = \frac{d}{2\varepsilon_0 S}\left[Q(Q-q) + \frac{q^2}{4}\right] = \frac{d}{2\varepsilon_0 S}\left(Q – \frac q2\right)^2$$

$$v = \left(Q – \frac q2\right)\sqrt{\frac{d}{m\,\varepsilon_0 S}}$$

where $m$ is the mass of the plate $\gamma$ (the book does not name it).

Question 29

There is another useful system of units, besides the SI/mks A system, called the cgs (centimetre-gram-second) system. In this system Coulomb’s law is given by $\vec F = \dfrac{Qq}{r^2}\hat r$, where the distance $r$ is measured in cm ($= 10^{-2}$ m), $F$ in dynes ($= 10^{-5}$ N) and the charges in electrostatic units (es units), where $1$ es unit of charge $= \dfrac{1}{[3]}\times 10^{-9}$ C. The number [3] actually arises from the speed of light in vacuum which is now taken to be exactly given by $c = 2.99792458\times10^{8}$ m/s. An approximate value of $c$ then is $c = [3]\times 10^{8}$ m/s. (i) Show that the Coulomb law in cgs units yields $1$ esu of charge $= 1\,(\text{dyne})^{1/2}$ cm. Obtain the dimensions of units of charge in terms of mass $M$, length $L$ and time $T$. Show that it is given in terms of fractional powers of $M$ and $L$. (ii) Write $1$ esu of charge $= x$ C, where $x$ is a dimensionless number. Show that this gives $\dfrac{1}{4\pi\varepsilon_0} = \dfrac{10^{-9}}{x^2}\ \dfrac{\text{N m}^2}{\text{C}^2}$. With $x = \dfrac{1}{[3]}\times10^{-9}$, we have $\dfrac{1}{4\pi\varepsilon_0} = [3]^2\times10^{9}\ \dfrac{\text{N m}^2}{\text{C}^2}$, or $\dfrac{1}{4\pi\varepsilon_0} = (2.99792458)^2\times10^{9}\ \dfrac{\text{N m}^2}{\text{C}^2}$ (exactly).

Solution.

(i) Put two equal charges of $1$ esu at a distance $1$ cm. Coulomb’s law in cgs units gives $F = \dfrac{(1)(1)}{1^2} = 1$ dyne. So the unit of charge satisfies

$$(1\ \text{esu})^2 = 1\ \text{dyne}\times(1\ \text{cm})^2 \;\Rightarrow\; 1\ \text{esu} = 1\ (\text{dyne})^{1/2}\ \text{cm}$$

For dimensions, $Q^2 = F r^2$ with $[F] = MLT^{-2}$ and $[r] = L$:

$$[Q] = (MLT^{-2})^{1/2}\,L = M^{1/2}L^{3/2}T^{-1}$$

The powers of $M$ and $L$ are fractional ($\tfrac12$ and $\tfrac32$), as required.

(ii) Take two charges of $x$ C each placed $1$ cm $= 10^{-2}$ m apart. They exert $1$ dyne $= 10^{-5}$ N on each other (this is what $1$ esu each means). In SI,

$$10^{-5} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{x^2}{(10^{-2})^2} \;\Rightarrow\; \frac{1}{4\pi\varepsilon_0} = \frac{10^{-9}}{x^2}\ \frac{\text{N m}^2}{\text{C}^2}$$

Substituting $x = \dfrac{10^{-9}}{[3]}$:

$$\frac{1}{4\pi\varepsilon_0} = 10^{-9}\,\frac{[3]^2}{10^{-18}} = [3]^2\times10^{9}\ \frac{\text{N m}^2}{\text{C}^2}$$

Since $c$ is exactly $2.99792458\times10^8$ m/s, [3] stands for $2.99792458$ here and

$$\frac{1}{4\pi\varepsilon_0} = (2.99792458)^2\times 10^{9} = 8.98755\times10^{9}\ \frac{\text{N m}^2}{\text{C}^2}$$

Question 30

Two charges $-q$ each are fixed separated by distance $2d$. A third charge $q$ of mass $m$ placed at the mid-point is displaced slightly by $x$ ($x \ll d$) perpendicular to the line joining the two fixed charged as shown in Fig. 1.14. Show that $q$ will perform simple harmonic oscillation of time period

$$T = \left[\frac{8\pi^3\varepsilon_0\,m\,d^3}{q^2}\right]^{1/2}$$

Solution. After the displacement, the charge $q$ is at distance $\sqrt{d^2+x^2}$ from each fixed charge. Each $-q$ attracts it with force $\dfrac{q^2}{4\pi\varepsilon_0(d^2+x^2)}$ along the line joining them, which makes angle $\theta$ with the perpendicular, where $\sin\theta = \dfrac{x}{\sqrt{d^2+x^2}}$. The components along the line joining the two fixed charges cancel by symmetry, and the components back towards the mid-point add:

$$F = -2\cdot\frac{q^2}{4\pi\varepsilon_0(d^2+x^2)}\cdot\frac{x}{\sqrt{d^2+x^2}} = -\frac{q^2\,x}{2\pi\varepsilon_0(d^2+x^2)^{3/2}}$$

For $x \ll d$, $(d^2+x^2)^{3/2} \approx d^3$, so

$$F \approx -\frac{q^2}{2\pi\varepsilon_0 d^3}\,x$$

This is a restoring force proportional to the displacement, so the motion is simple harmonic with $\omega^2 = \dfrac{q^2}{2\pi\varepsilon_0 m d^3}$. Then

$$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{2\pi\varepsilon_0 m d^3}{q^2}} = \left[\frac{8\pi^3\varepsilon_0\,m\,d^3}{q^2}\right]^{1/2}$$

Question 31

Total charge $-Q$ is uniformly spread along length of a ring of radius $R$. A small test charge $+q$ of mass $m$ is kept at the centre of the ring and is given a gentle push along the axis of the ring. (a) Show that the particle executes a simple harmonic oscillation. (b) Obtain its time period.

Solution. (a) Let the test charge be at distance $z$ from the centre along the axis. Every element of the ring is at the same distance $\sqrt{R^2+z^2}$ from it. The sideways components of the attraction cancel by symmetry, and only the components along the axis, pointing back towards the centre, add up. The result is the field on the axis of a charged ring:

$$F = -\frac{Q\,q\,z}{4\pi\varepsilon_0\,(R^2+z^2)^{3/2}}$$

For a small push, $z \ll R$, so $(R^2+z^2)^{3/2} \approx R^3$:

$$F \approx -\frac{Q\,q}{4\pi\varepsilon_0 R^3}\,z$$

The force opposes the displacement and is proportional to it, so the motion is simple harmonic.

(b) Comparing with $F = -m\omega^2 z$:

$$\omega^2 = \frac{Qq}{4\pi\varepsilon_0 m R^3}, \qquad T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{4\pi\varepsilon_0\,m\,R^3}{Qq}}$$

Common mistakes

• Saying zero charge means zero field, in Question 17. Gauss’s law fixes only the net flux. The field on the surface is generally not zero for a dipole inside it.
• Dividing the cube flux wrongly in Question 19. Count the cubes that would meet at the charge: $8$ at a corner, $4$ at an edge, $2$ at a face. Putting the charge at the mid-point of B and C does not make a new case, because that point is on a face.
• Using the wrong sign of the third charge’s position in Question 23. The balance point lies outside the pair, on the side of the smaller charge; between the two opposite charges both fields point the same way and cannot cancel.
• Treating the cube of CsCl as having a net field in Question 22(ii). The trick is subtraction: a full set gives zero, so the missing ion leaves the reverse of its own contribution.
• Forgetting that a plate’s field does not act on itself, in Question 28. The field on $\gamma$ comes from $\alpha$ and $\beta$ only; and each plate’s field is $\text{charge}/2\varepsilon_0 S$, not $\text{charge}/\varepsilon_0 S$.
• Losing the symmetry argument in Questions 30 and 31. The sideways components cancel exactly; only the restoring component, which is proportional to the displacement for small $x$ or $z$, survives. Forgetting the factor $2$ in Question 30 gives a period too long by $\sqrt2$.
• Mixing up the cgs relation in Question 29. $1$ esu is $\tfrac{1}{[3]}\times10^{-9}$ C, so a larger [3] means a smaller esu. The $c^2$ appearing in $1/4\pi\varepsilon_0$ comes from squaring that factor.

Practise next

• Exemplar Multiple Choice Questions — the thirteen questions on the same relations, including the two places where this unit’s printed key is wrong.
• NCERT textbook Chapter 1 — Electric Charges and Fields for the end-of-chapter problems on Coulomb’s law, dipoles and Gauss’s law.
• Exemplar Electrostatic Potential and Capacitance — the next unit, which turns the same fields into potentials and energies.