NCERT Class 12 Physics — Electric Charges and Fields, Chapter 1 Exercises. All 23 questions solved.

The exercises at the end of Chapter 1 fall into three groups. Questions 1.1 to 1.6, 1.11 and 1.12 are about charge itself and Coulomb’s law. Questions 1.7 to 1.10 and 1.13 are about the electric field — its direction, its field lines and what it does to a dipole or a moving charge. Questions 1.14 to 1.23 are about electric flux and Gauss’s law. Almost all the arithmetic runs on three results:

• Coulomb’s law — $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$, with $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^9\ \text{N m}^2\,\text{C}^{-2}$.
• Electric flux through a flat area — $\Phi = \vec E\cdot\Delta\vec S = E\,\Delta S\cos\theta$, where $\theta$ is the angle between the field and the normal to the area.
• Gauss’s law — the flux out of any closed surface is $\Phi = q_{\text{enclosed}}/\varepsilon_0$, with $\varepsilon_0 = 8.854\times10^{-12}\ \text{C}^2\,\text{N}^{-1}\,\text{m}^{-2}$.

Every distance given in centimetres is converted to metres before it goes into a formula.

Key insight. Look for the symmetry before you calculate. In this exercise it settles most answers before any arithmetic: equal charges at opposite corners cancel (1.6), a uniform field sends as much flux out of a closed box as in (1.15), a charge at the centre of a cube shares its flux equally among the six faces (1.17), and outside two oppositely charged plates the two fields cancel (1.23). Once the symmetry is found, what is left is usually one line of Coulomb’s law or Gauss’s law.

Question 1.1

What is the force between two small charged spheres having charges of $2\times10^{-7}\ \text{C}$ and $3\times10^{-7}\ \text{C}$ placed $30\ \text{cm}$ apart in air?

Solution. The spheres are small compared with their separation, so they behave as point charges and Coulomb’s law applies directly. With $r = 0.30\ \text{m}$:

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} = \frac{(9\times10^{9})(2\times10^{-7})(3\times10^{-7})}{(0.30)^2} = \frac{5.4\times10^{-4}}{0.09} = 6\times10^{-3}\ \text{N}$$

Both charges are positive, so the force pushes the spheres apart.

Question 1.2

The electrostatic force on a small sphere of charge $0.4\ \mu\text{C}$ due to another small sphere of charge $-0.8\ \mu\text{C}$ in air is $0.2\ \text{N}$. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

Question 1.2 (a)

Solution. Coulomb’s law can be turned round to give the distance. Only the magnitudes matter here, since the signs tell us the direction and not the size:

$$r^2 = \frac{1}{4\pi\varepsilon_0}\frac{|q_1 q_2|}{F} = \frac{(9\times10^9)(0.4\times10^{-6})(0.8\times10^{-6})}{0.2} = 1.44\times10^{-2}\ \text{m}^2$$

so $r = 0.12\ \text{m}$.

Question 1.2 (b)

Solution. The two forces between a pair of charges are equal and opposite, as Newton’s third law requires, and Coulomb’s law is symmetric in $q_1$ and $q_2$. So the fact that one charge is twice the other changes nothing. The charges have opposite signs, so the force is attractive: the second sphere is pulled towards the first.

Question 1.3

Check that the ratio $ke^2/Gm_em_p$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Solution. Dimensions. $k$ has units $\text{N m}^2\,\text{C}^{-2}$, so $ke^2$ has units $\text{N m}^2$. $G$ has units $\text{N m}^2\,\text{kg}^{-2}$, so $Gm_em_p$ also has units $\text{N m}^2$. The units cancel and the ratio is a pure number.

Value. With $k = 9\times10^9$, $e = 1.6\times10^{-19}\ \text{C}$, $G = 6.67\times10^{-11}$, $m_e = 9.1\times10^{-31}\ \text{kg}$ and $m_p = 1.67\times10^{-27}\ \text{kg}$:

$$\frac{ke^2}{Gm_em_p} = \frac{(9\times10^9)(1.6\times10^{-19})^2}{(6.67\times10^{-11})(9.1\times10^{-31})(1.67\times10^{-27})} = \frac{2.30\times10^{-28}}{1.01\times10^{-67}} \approx 2.3\times10^{39}$$

Meaning. $ke^2/r^2$ is the electric force between an electron and a proton a distance $r$ apart, and $Gm_em_p/r^2$ is the gravitational force between them. The $r^2$ cancels, so the ratio is the same at every distance: the electric attraction is about $10^{39}$ times the gravitational one. That is why gravity plays no part in holding an atom together.

Note on the book’s answer. The answer key prints $2.4\times10^{39}$. Working it through with the constants above, or with more precise ones, gives $2.27\times10^{39}$, so $2.3\times10^{39}$ is the value to quote. The conclusion, a ratio of the order of $10^{39}$, is the same either way.

Question 1.4

(a) Explain the meaning of the statement ‘electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic, i.e., large scale charges?

Question 1.4 (a)

Solution. Charge comes only in whole-number multiples of a smallest unit, the magnitude of the electron’s charge, $e = 1.6\times10^{-19}\ \text{C}$. Charge is transferred by moving whole electrons, so any charge a body carries is $q = ne$ with $n$ a positive or negative integer; a body cannot carry $1.5e$ or $0.3e$.

Question 1.4 (b)

Solution. At everyday scales the step size is far too small to notice. A modest charge of $1\ \mu\text{C}$ is $10^{-6}/(1.6\times10^{-19}) \approx 6\times10^{12}$ electronic charges. Adding or removing one electron changes it by about one part in $10^{13}$, far below anything a measurement can resolve, so charge behaves as though it could take any value.

Question 1.5

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Solution. Before rubbing, the rod and the cloth are both neutral, so the total charge is zero. Rubbing does not create charge; it moves electrons from the glass to the silk. The glass is left positive and the silk becomes negative by exactly the same amount, so the total is still zero. The same is true of every pair of bodies rubbed together: the two always acquire equal and opposite charges.

Question 1.6

Four point charges $q_A = 2\ \mu\text{C}$, $q_B = -5\ \mu\text{C}$, $q_C = 2\ \mu\text{C}$ and $q_D = -5\ \mu\text{C}$ are located at the corners of a square $ABCD$ of side $10\ \text{cm}$. What is the force on a charge of $1\ \mu\text{C}$ placed at the centre of the square?

Solution. The centre $O$ is the same distance, $5\sqrt2\ \text{cm}$, from all four corners. $A$ and $C$ are opposite corners carrying equal charges, so they push the charge at $O$ with equal forces in exactly opposite directions, and those two forces cancel. $B$ and $D$ are the other pair of opposite corners, also carrying equal charges, and their two attractions cancel in the same way. Nothing is left over, so the individual forces never need to be calculated.

Question 1.7

(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point.

Question 1.7 (a)

Solution. A field line is drawn so that its tangent at each point gives the direction of the field there. In a region with no charges the electrostatic field exists at every point and changes smoothly from one point to the next. A gap in a field line would mean the field suddenly vanishing or jumping at that point, which a smoothly varying field does not do. Field lines can begin only on positive charges and end only on negative ones.

Question 1.7 (b)

Solution. The field at a point is a single vector — the sum of the contributions of all the charges — so it has exactly one direction there. If two field lines crossed, the crossing point would have two tangents, meaning two field directions at once, which is impossible.

Question 1.8

Two point charges $q_A = 3\ \mu\text{C}$ and $q_B = -3\ \mu\text{C}$ are located $20\ \text{cm}$ apart in vacuum. (a) What is the electric field at the midpoint $O$ of the line $AB$ joining the two charges? (b) If a negative test charge of magnitude $1.5\times10^{-9}\ \text{C}$ is placed at this point, what is the force experienced by the test charge?

Question 1.8 (a)

Solution. $O$ is $0.10\ \text{m}$ from each charge, and the charges have equal magnitude, so each produces a field of the same size at $O$:

$$E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} = \frac{(9\times10^9)(3\times10^{-6})}{(0.10)^2} = 2.7\times10^{6}\ \text{N C}^{-1}$$

The direction is where it is easy to go wrong. The field of the positive charge at $A$ points away from $A$, that is, towards $B$. The field of the negative charge at $B$ points towards $B$. Both point the same way, from $A$ towards $B$, so they add rather than cancel:

$$E_O = 2\times2.7\times10^6 = 5.4\times10^6\ \text{N C}^{-1}$$

Question 1.8 (b)

Solution. The force on a charge in a field is $F = qE$, and on a negative charge it points opposite to the field:

$$F = (1.5\times10^{-9})(5.4\times10^6) = 8.1\times10^{-3}\ \text{N}$$

Question 1.9

A system has two charges $q_A = 2.5\times10^{-7}\ \text{C}$ and $q_B = -2.5\times10^{-7}\ \text{C}$ located at points $A: (0, 0, -15\ \text{cm})$ and $B: (0, 0, +15\ \text{cm})$, respectively. What are the total charge and electric dipole moment of the system?

Solution. The total charge is simply the sum, $2.5\times10^{-7} – 2.5\times10^{-7} = 0$.

Equal and opposite charges a distance $2a$ apart form a dipole of moment $p = q\times2a$. Here $2a = 30\ \text{cm} = 0.30\ \text{m}$:

$$p = (2.5\times10^{-7})(0.30) = 7.5\times10^{-8}\ \text{C m}$$

The dipole moment points from the negative charge to the positive one. The negative charge is at $B$, on the positive $z$-axis, and the positive charge is at $A$, on the negative $z$-axis, so $\vec p$ points in the $-z$ direction. Adding up $\sum q_i\vec r_i$ confirms it: $(2.5\times10^{-7})(-0.15\,\hat k) + (-2.5\times10^{-7})(0.15\,\hat k) = -7.5\times10^{-8}\,\hat k\ \text{C m}$.

Note on the book’s answer. The key says “along $z$-axis”, which names the line but not the sense; the dipole points towards negative $z$.

Question 1.10

An electric dipole with dipole moment $4\times10^{-9}\ \text{C m}$ is aligned at $30^\circ$ with the direction of a uniform electric field of magnitude $5\times10^4\ \text{N C}^{-1}$. Calculate the magnitude of the torque acting on the dipole.

Solution. In a uniform field the forces on the two charges of a dipole are equal and opposite, so there is no net force. They act along different lines, though, and so form a couple whose torque is $\tau = pE\sin\theta$:

$$\tau = (4\times10^{-9})(5\times10^{4})\sin30^\circ = (2\times10^{-4})\times\tfrac12 = 10^{-4}\ \text{N m}$$

The torque turns the dipole towards alignment with the field.

Question 1.11

A polythene piece rubbed with wool is found to have a negative charge of $3\times10^{-7}\ \text{C}$. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene?

Question 1.11 (a)

Solution. The polythene has gained electrons, each carrying charge $e$, so their number is

$$n = \frac{q}{e} = \frac{3\times10^{-7}}{1.6\times10^{-19}} = 1.875\times10^{12} \approx 2\times10^{12}$$

Since the polythene became negative, the electrons moved from the wool to the polythene.

Question 1.11 (b)

Solution. Yes, because electrons have mass. The mass carried across is

$$m = n\,m_e = (1.875\times10^{12})(9.1\times10^{-31}) \approx 1.7\times10^{-18}\ \text{kg}$$

or $2\times10^{-18}\ \text{kg}$ if $n$ is rounded to $2\times10^{12}$ first, as the book’s key does. Either way it is utterly negligible.

Question 1.12

(a) Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50\ \text{cm}$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times10^{-7}\ \text{C}$? The radii of $A$ and $B$ are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Question 1.12 (a)

Solution. With radii negligible, the spheres act as point charges at their centres, $r = 0.50\ \text{m}$ apart:

$$F = \frac{(9\times10^9)(6.5\times10^{-7})^2}{(0.50)^2} = \frac{3.80\times10^{-3}}{0.25} = 1.52\times10^{-2}\ \text{N}$$

Question 1.12 (b)

Solution. Scaling is quicker than recalculating. Doubling each charge multiplies $q_1q_2$ by $4$, and halving the distance multiplies $1/r^2$ by another $4$, so the force becomes $16$ times as large:

$$F’ = 16\times1.52\times10^{-2} = 0.24\ \text{N}$$

Question 1.13

Figure 1.30 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Solution. The field between the plates points from the positive plate to the negative one, downwards in the figure. A positive charge is pushed along the field and a negative charge against it. Particles 1 and 2 bend upwards, towards the positive plate, so they are negative; particle 3 bends downwards, towards the negative plate, so it is positive.

For the charge to mass ratio, take a particle entering with horizontal speed $v$. Its vertical acceleration is $qE/m$, and after travelling a horizontal distance $x$, which takes time $x/v$, it has been deflected by

$$y = \frac12\,\frac{qE}{m}\left(\frac{x}{v}\right)^2$$

For the same field, the same distance $x$ and, as the question intends, the same entry speed, the deflection is proportional to $q/m$. Particle 3 is deflected the most.

Question 1.14

Consider a uniform electric field $\vec E = 3\times10^3\ \hat{\imath}\ \text{N/C}$. (a) What is the flux of this field through a square of $10\ \text{cm}$ on a side whose plane is parallel to the $yz$ plane? (b) What is the flux through the same square if the normal to its plane makes a $60^\circ$ angle with the $x$-axis?

Question 1.14 (a)

Solution. A plane parallel to the $yz$ plane has its normal along the $x$-axis, the same direction as the field, so $\theta = 0$. The area is $(0.10)^2 = 1.0\times10^{-2}\ \text{m}^2$:

$$\Phi = E\,\Delta S\cos0^\circ = (3\times10^3)(1.0\times10^{-2}) = 30\ \text{N m}^2/\text{C}$$

Question 1.14 (b)

Solution. The angle in the flux formula is the one between the field and the normal, which is exactly what the question gives: $\theta = 60^\circ$.

$$\Phi = 30\cos60^\circ = 15\ \text{N m}^2/\text{C}$$

Question 1.15

What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side $20\ \text{cm}$ oriented so that its faces are parallel to the coordinate planes?

Solution. The field is along $x$. On the face where it enters, the outward normal points along $-x$ and the flux is $-EA$; on the opposite face, where it leaves, the flux is $+EA$. The other four faces are parallel to the field, so no field lines cross them. The total is zero. Gauss’s law gives the same result directly: the cube encloses no charge, so the net flux out of it must be zero.

Question 1.16

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0\times10^3\ \text{N m}^2/\text{C}$. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

Question 1.16 (a)

Solution. Gauss’s law, $\Phi = q/\varepsilon_0$, holds for a closed surface of any shape, so the charge follows at once:

$$q = \varepsilon_0\Phi = (8.854\times10^{-12})(8.0\times10^3) = 7.1\times10^{-8}\ \text{C}$$

Question 1.16 (b)

Solution. No. Zero flux tells us only that the net charge inside is zero. The box could hold equal positive and negative charges whose contributions to the flux cancel.

Question 1.17

A point charge $+10\ \mu\text{C}$ is a distance $5\ \text{cm}$ directly above the centre of a square of side $10\ \text{cm}$, as shown in Fig. 1.31. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge $10\ \text{cm}$.)

Solution. Working out the flux through the square directly would need an integral, because the field changes in size and direction across it. The hint avoids that. A cube of edge $10\ \text{cm}$ built on the square has its centre $5\ \text{cm}$ above the centre of the square — exactly where the charge is. By Gauss’s law the total flux out of that cube is $q/\varepsilon_0$, and since the charge is at the centre, all six faces are identical as seen from it and share the flux equally:

$$\Phi_{\text{square}} = \frac{q}{6\varepsilon_0} = \frac{10\times10^{-6}}{6\times8.854\times10^{-12}} = 1.88\times10^5\ \text{N m}^2/\text{C}$$

Note on the book’s answer. The answer key prints $2.2\times10^5$ here and $1.9\times10^5$ for Question 1.18. The two have been printed against the wrong questions: $1.9\times10^5$ is this question’s $q/6\varepsilon_0$, and Question 1.18’s $q/\varepsilon_0$ works out to $2.26\times10^5$.

Question 1.18

A point charge of $2.0\ \mu\text{C}$ is at the centre of a cubic Gaussian surface $9.0\ \text{cm}$ on edge. What is the net electric flux through the surface?

Solution. The flux out of a closed surface depends only on the charge inside it, not on the surface’s shape or size, so the $9.0\ \text{cm}$ plays no part:

$$\Phi = \frac{q}{\varepsilon_0} = \frac{2.0\times10^{-6}}{8.854\times10^{-12}} = 2.26\times10^5\ \text{N m}^2/\text{C}$$

Note on the book’s answer. The key prints $1.9\times10^5$, which is the answer to Question 1.17; see the note there.

Question 1.19

A point charge causes an electric flux of $-1.0\times10^3\ \text{N m}^2/\text{C}$ to pass through a spherical Gaussian surface of $10.0\ \text{cm}$ radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Question 1.19 (a)

Solution. Doubling the radius makes the field at the surface a quarter as strong but the area four times as large, so the flux is unchanged. That is Gauss’s law again: the enclosed charge is the same, so the flux is the same.

Question 1.19 (b)

Solution. The negative flux means the field lines point inwards, so the charge is negative:

$$q = \varepsilon_0\Phi = (8.854\times10^{-12})(-1.0\times10^3) = -8.85\times10^{-9}\ \text{C}$$

Question 1.20

A conducting sphere of radius $10\ \text{cm}$ has an unknown charge. If the electric field $20\ \text{cm}$ from the centre of the sphere is $1.5\times10^3\ \text{N/C}$ and points radially inward, what is the net charge on the sphere?

Solution. Outside a charged conducting sphere the field is the same as if all its charge were at the centre, so $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ with $r$ measured from the centre — here $0.20\ \text{m}$. The sphere’s own radius does not enter.

$$|q| = \frac{Er^2}{9\times10^9} = \frac{(1.5\times10^3)(0.20)^2}{9\times10^9} = 6.67\times10^{-9}\ \text{C}$$

A field pointing inwards means the charge is negative.

Question 1.21

A uniformly charged conducting sphere of $2.4\ \text{m}$ diameter has a surface charge density of $80.0\ \mu\text{C/m}^2$. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Question 1.21 (a)

Solution. Charge is surface density times area, and the radius is half the diameter, $1.2\ \text{m}$:

$$q = \sigma\cdot4\pi r^2 = (80.0\times10^{-6})(4\pi)(1.2)^2 = 1.45\times10^{-3}\ \text{C}$$

Question 1.21 (b)

Solution. All of that charge is enclosed by the sphere’s surface, so Gauss’s law gives the flux:

$$\Phi = \frac{q}{\varepsilon_0} = \frac{1.45\times10^{-3}}{8.854\times10^{-12}} = 1.6\times10^8\ \text{N m}^2/\text{C}$$

Question 1.22

An infinite line charge produces a field of $9\times10^4\ \text{N/C}$ at a distance of $2\ \text{cm}$. Calculate the linear charge density.

Solution. Gauss’s law applied to a cylinder around the line gives the field of an infinite line charge:

$$E = \frac{\lambda}{2\pi\varepsilon_0 r} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2\lambda}{r}$$

Solving for $\lambda$, with $r = 0.02\ \text{m}$:

$$\lambda = \frac{Er}{2}\cdot4\pi\varepsilon_0 = \frac{(9\times10^4)(0.02)}{2\times9\times10^9} = 1.0\times10^{-7}\ \text{C/m}$$

Note on the book’s answer. The key prints $10\ \mu\text{C/m}$, which is $10^{-5}\ \text{C/m}$ — a hundred times the value above. Putting $10\ \mu\text{C/m}$ back into the formula gives a field of $9\times10^6\ \text{N/C}$ at $2\ \text{cm}$, not the $9\times10^4\ \text{N/C}$ in the question, so the printed answer is a misprint. The charge density is $0.10\ \mu\text{C/m}$.

Question 1.23

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 $17.0\times10^{-22}\ \text{C/m}^2$. What is $\vec E$: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

Solution. Each charged sheet produces a uniform field of magnitude $\sigma/2\varepsilon_0$, pointing away from it if it is positive and towards it if it is negative. Here one sheet carries $+\sigma$ and the other $-\sigma$.

Outside either plate, the field of the positive plate points away from it and the field of the negative plate points towards it, and on the outside those are opposite directions. The two are equal in size, so they cancel.

Between the plates, both fields point from the positive plate to the negative one, so they add:

$$E = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} = \frac{17.0\times10^{-22}}{8.854\times10^{-12}} = 1.92\times10^{-10}\ \text{N/C}$$

Note on the book’s answer. The key gives $1.9\ \text{N/C}$ for (c). The digits are right but the power of ten has dropped out: with $\sigma = 17.0\times10^{-22}\ \text{C/m}^2$, the field between the plates is $1.9\times10^{-10}\ \text{N/C}$.

Common mistakes

• Question 1.2(b): expecting a different force on the second sphere. Because one charge is twice the other, it feels as though the forces should differ. They cannot: Coulomb’s law is symmetric in the two charges, and the forces are a Newton’s third law pair.
• Question 1.8(a): subtracting the two fields at the midpoint. The charges are opposite, so students cancel their fields. But at a point between a positive and a negative charge, both fields point towards the negative charge, and they add to $5.4\times10^6\ \text{N C}^{-1}$.
• Question 1.9: giving the dipole moment the wrong sense. The dipole moment points from the negative charge to the positive one, which here is towards $-z$. It is easy to miss when the book’s key says only “along $z$-axis”.
• Question 1.12(b): scaling by 4 instead of 16. Doubling each charge and halving the distance each multiply the force by 4, so together they multiply it by 16. Students often apply only one of the two, or let them cancel.
• Question 1.14(b): using $\sin60^\circ$. The angle in $\Phi = E\,\Delta S\cos\theta$ is between the field and the normal to the surface, not the surface itself. The question gives exactly that angle, so it is $\cos60^\circ$.
• Question 1.17: using $q/\varepsilon_0$ for the square. Gauss’s law gives the flux through a closed surface. The square is one face of six, so it gets $q/6\varepsilon_0$.
• Question 1.20: putting in the radius of the sphere. The field is measured $20\ \text{cm}$ from the centre, and that is the $r$ in the formula; the sphere’s $10\ \text{cm}$ radius plays no part outside it.
• Question 1.22: using the point-charge formula. The field of a line charge is $\lambda/2\pi\varepsilon_0 r$. It falls off as $1/r$, not $1/r^2$, and carries a factor of 2 compared with $\frac{1}{4\pi\varepsilon_0}\frac{\lambda}{r}$.

Practise next

• Chapter 2, Electrostatic Potential and Capacitance — the same charges and fields, described through energy rather than force.