Exercises — Electric Charges and Fields

1. 12 marksNCERT Exercises, Chapter 1

   What is the force between two small charged spheres having charges of $2\times 10^{-7}$ C and $3\times 10^{-7}$ C placed 30 cm apart in air?

   Hint. Apply Coulomb's law directly, converting 30 cm to metres first.

   Coulomb's law gives the force between two point charges as $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$.

   Here $q_1=2\times 10^{-7}$ C, $q_2=3\times 10^{-7}$ C and $r=30$ cm $=0.30$ m.

   $F=\frac{9\times 10^9\times 2\times 10^{-7}\times 3\times 10^{-7}}{(0.30{)}^2}=\frac{5.4\times 10^{-4}}{0.09}$

   so $F=6\times 10^{-3}$ N.

   Both charges are positive, so the force is repulsive — stating the direction is part of the answer, since force is a vector.

   ✦ F = 6 x 10^-3 N, repulsive

2. 23 marksNCERT Exercises, Chapter 1

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

   Hint. Rearrange Coulomb's law for r in part (a). Part (b) needs no calculation at all.

   Rearranging Coulomb's law for the separation is the whole of part (a).

   (a) From $F=\frac{k{q}_1{q}_2}{r^2}$ we get $r^2=\frac{k{q}_1{q}_2}{F}$.

   $r^2=\frac{9\times 10^9\times 0.4\times 10^{-6}\times 0.8\times 10^{-6}}{0.2}=\frac{2.88\times 10^{-3}}{0.2}=0.0144$

   so $r=0.12$ m $=12$ cm.

   (b) By Newton's third law the two spheres exert equal and opposite forces on each other, so the force on the second sphere is also $0.2$ N. Since the charges have opposite signs, both forces are attractive.

   No fresh calculation is needed for (b) — recognising that is the point of the question.

   ✦ (a) r = 0.12 m = 12 cm (b) 0.2 N, attractive (Newton's third law)

3. 33 marksNCERT Exercises, Chapter 1

   Check that the ratio $k{e}^2/G{m}_e{m}_p$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

   Hint. Write the dimensions of the electrostatic and gravitational force expressions and note that both are forces.

   The two products are the numerators of the Coulomb and gravitational force laws, both divided by the same $r^2$.

   Since $F_{\text{electric}}=\frac{k{e}^2}{r^2}$ and $F_{\text{grav}}=\frac{G{m}_e{m}_p}{r^2}$ are both forces, the quantities $k{e}^2$ and $G{m}_e{m}_p$ have identical dimensions, so their ratio is dimensionless.

   Substituting $k=9\times 10^9$, $e=1.6\times 10^{-19}$ C, $G=6.67\times 10^{-11}$, $m_e=9.11\times 10^{-31}$ kg and $m_p=1.67\times 10^{-27}$ kg:

   $\frac{k{e}^2}{G{m}_e{m}_p}=\frac{9\times 10^9\times (1.6\times 10^{-19}{)}^2}{6.67\times 10^{-11}\times 9.11\times 10^{-31}\times 1.67\times 10^{-27}}=\frac{2.30\times 10^{-28}}{1.01\times 10^{-67}}$

   which gives approximately $2.3\times 10^{39}$.

   This is the ratio of the electrostatic force to the gravitational force between an electron and a proton. Its enormous size is why gravity is completely ignored in atomic physics.

   ✦ Dimensionless; approximately 2.3 x 10^39, the ratio of electric to gravitational force between an electron and a proton

4. 43 marksNCERT Exercises, Chapter 1

   (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?

   Hint. Think about the size of e compared with the charges handled in a laboratory.

   (a) Quantisation means that the charge on any body is always an integral multiple of the elementary charge $e=1.6\times 10^{-19}$ C. That is, $q=ne$ where $n$ is an integer, positive or negative.

   Charge cannot take arbitrary values — a body may carry $e$, $2e$ or $-3e$, but never a fraction such as $0.5e$. This is because charge is transferred only by whole electrons.

   (b) At the macroscopic scale the charges encountered are enormous compared with $e$. A charge of $1\mu$C already corresponds to about $10^{13}$ electrons.

   Adding or removing one electron changes such a charge by roughly one part in $10^{13}$, which no measurement can detect. The steps are so fine relative to the total that charge appears continuous, so quantisation can safely be ignored.

   ✦ (a) Charge is always an integral multiple of e = 1.6 x 10^-19 C (b) Because macroscopic charges are around 10^13 times e, the steps are undetectably small and charge appears continuous

5. 53 marksNCERT Exercises, Chapter 1

   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.

   Hint. Consider the total charge of the two bodies taken together, before and after rubbing.

   Rubbing does not create charge; it only transfers electrons from one body to the other.

   When glass is rubbed with silk, electrons move from the glass to the silk. The glass is left with a deficit of electrons and becomes positively charged, while the silk gains exactly those electrons and becomes negatively charged.

   Crucially, the two charges are equal in magnitude and opposite in sign.

   So if the pair is treated as one isolated system, the total charge before rubbing was zero, and after rubbing it is $+q+(-q)=0$ once more.

   Since the total charge of the isolated system is unchanged, the observation is entirely consistent with the law of conservation of charge. The same reasoning applies to every other pair of bodies showing this effect.

   ✦ Electrons are only transferred, not created: the two bodies acquire equal and opposite charges, so the total charge of the isolated system stays zero

6. 63 marksNCERT Exercises, Chapter 1

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

   Hint. Identify which charges sit diagonally opposite each other before computing anything.

   Look at the geometry before reaching for Coulomb's law, because the symmetry settles the whole question.

   In square ABCD the diagonally opposite pairs are A and C, and B and D. The centre is equidistant from all four corners.

   Charges $q_A$ and $q_C$ are equal at $2\mu$C each and sit at opposite ends of a diagonal, so they exert forces on the centre charge that are equal in magnitude and exactly opposite in direction. These cancel.

   Similarly $q_B$ and $q_D$ are equal at $-5\mu$C each and also diagonally opposite, so their forces cancel too.

   Since both pairs cancel independently, the resultant force is zero.

   No numerical work is required, and attempting to compute the four individual forces wastes time.

   ✦ Zero — the diagonally opposite equal charges produce forces that cancel in pairs

7. 73 marksNCERT Exercises, Chapter 1

   (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.

   Hint. In each part, ask what a break or a crossing would imply about the field at that point.

   (a) A field line traces the path a small positive test charge would follow. The electric field exists at every point of space around a charge, so the test charge experiences a force everywhere along its path and moves continuously.

   A sudden break would mean the field abruptly ceased to exist at that point, which does not happen except where the line terminates on a charge. Hence field lines are continuous curves.

   (b) The tangent to a field line at any point gives the direction of the electric field there.

   If two lines crossed, two different tangents could be drawn at the point of intersection, implying two different directions for the field at the same location.

   That is impossible, since the force on a test charge placed there is a single well-defined vector with one direction. Therefore field lines never cross.

   ✦ (a) The field exists everywhere along the path, so the test charge moves continuously (b) A crossing would give two directions for the field at one point, which is impossible

8. 84 marksNCERT Exercises, Chapter 1

   Two point charges $q_A=3\mu$C and $q_B=-3\mu$C are located 20 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 $1.5\times 10^{-9}$ C is placed at this point, what is the force experienced by the test charge?

   Hint. At the midpoint, check whether the two fields point the same way or opposite ways before adding.

   The midpoint lies 10 cm from each charge, so each contributes a field of the same magnitude.

   (a) $E=\frac{kq}{r^2}=\frac{9\times 10^9\times 3\times 10^{-6}}{(0.10{)}^2}=2.7\times 10^6$ N/C from each charge.

   The field due to the positive charge at A points away from A, that is from A towards B. The field due to the negative charge at B points towards B, which is also from A towards B.

   Both contributions therefore point the same way and add:

   $E_{\text{total}}=2\times 2.7\times 10^6=5.4\times 10^6\text{N/C, directed from A to B}$

   (b) The force on the test charge is $F=|q|E=1.5\times 10^{-9}\times 5.4\times 10^6=8.1\times 10^{-3}$ N.

   Since the test charge is negative, the force is opposite to the field, so it is directed from B towards A.

   ✦ (a) 5.4 x 10^6 N/C directed from A to B (b) 8.1 x 10^-3 N directed from B to A

9. 93 marksNCERT Exercises, Chapter 1

   A system has two charges $q_A=2.5\times 10^{-7}$ C and $q_B=-2.5\times 10^{-7}$ C located at points A: (0, 0, -15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

   Hint. The separation is the full distance between the two charges, not the distance from the origin.

   Adding the charges algebraically gives the total, and the geometry gives the dipole moment.

   Total charge $=2.5\times 10^{-7}+(-2.5\times 10^{-7})=0$.

   The system is therefore an electric dipole — equal and opposite charges separated by a distance.

   The separation is the full distance from A to B, which is $15+15=30$ cm $=0.30$ m, not $15$ cm.

   $p=q\times 2a=2.5\times 10^{-7}\times 0.30=7.5\times 10^{-8}\text{C m}$

   By convention the dipole moment points from the negative charge to the positive charge, that is from B at $+z$ towards A at $-z$. So $\vec{p}$ is directed along the negative $z$-axis.

   ✦ Total charge = 0; p = 7.5 x 10^-8 C m directed from B to A, along the negative z-axis

10. 102 marksNCERT Exercises, Chapter 1

    An electric dipole with dipole moment $4\times 10^{-9}$ C m is aligned at $30^{\circ }$ with the direction of a uniform electric field of magnitude $5\times 10^4$ NC${}^{-1}$. Calculate the magnitude of the torque acting on the dipole.

    Hint. Use the magnitude form of the torque on a dipole in a uniform field.

    A dipole in a uniform field experiences no net force but does experience a torque, whose magnitude is $\tau =pE\sin \theta$.

    Here $p=4\times 10^{-9}$ C m, $E=5\times 10^4$ N/C and $\theta =30^{\circ }$, so $\sin \theta =0.5$.

    $\tau =4\times 10^{-9}\times 5\times 10^4\times 0.5=1\times 10^{-4}\text{N m}$

    Note that the torque vanishes when the dipole is aligned with the field ($\theta =0$) and is greatest when it is perpendicular to it, which is why $\sin \theta$ rather than $\cos \theta$ appears.

    ✦ Torque = 10^-4 N m

11. 113 marksNCERT Exercises, Chapter 1

    A polythene piece rubbed with wool is found to have a negative charge of $3\times 10^{-7}$ C. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene?

    Hint. Use quantisation of charge for (a), then multiply by the electron mass for (b).

    (a) Since charge is quantised, $q=ne$, so the number of electrons is:

    $n=\frac{q}{e}=\frac{3\times 10^{-7}}{1.6\times 10^{-19}}\approx 1.87\times 10^{12}$

    The polythene has become negative, meaning it gained electrons. So roughly $1.87\times 10^{12}$ electrons were transferred from the wool to the polythene.

    (b) Yes, because electrons carry mass. The mass transferred is:

    $\Delta m=n\times m_e=1.87\times 10^{12}\times 9.11\times 10^{-31}\approx 1.7\times 10^{-18}\text{kg}$

    The transfer is from wool to polythene, in the same direction as the electrons. This mass is utterly negligible compared with the mass of either object, which is why charging a body is never observed to change its weight.

    ✦ (a) About 1.87 x 10^12 electrons, transferred from wool to polythene (b) Yes, about 1.7 x 10^-18 kg, entirely negligible

12. 123 marksNCERT Exercises, Chapter 1

    (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is $6.5\times 10^{-7}$ 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?

    Hint. In (b), track how each change scales the force rather than recomputing from scratch.

    Because the radii are negligible the spheres behave as point charges, so Coulomb's law applies directly.

    (a) $F=\frac{k{q}^2}{r^2}=\frac{9\times 10^9\times (6.5\times 10^{-7}{)}^2}{(0.50{)}^2}=\frac{3.80\times 10^{-3}}{0.25}$

    so $F=1.52\times 10^{-2}$ N.

    (b) Doubling each charge multiplies the product $q_1{q}_2$ by $2\times 2=4$. Halving the distance divides $r^2$ by $4$, which multiplies the force by a further $4$.

    The net effect is a factor of $4\times 4=16$, giving:

    $F^{\prime }=16\times 1.52\times 10^{-2}=0.243\text{N}$

    Reasoning by scaling is faster than substituting again and makes the factor of 16 explicit.

    ✦ (a) F = 1.52 x 10^-2 N (b) F' = 0.243 N, which is 16 times larger

13. 133 marksNCERT Exercises, Chapter 1

    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?

    Hint. Deflection direction gives the sign; the amount of deflection over the same distance gives the charge-to-mass ratio.

    Two separate features of each track carry the information.

    Signs. A positive charge is deflected in the direction of the field, and a negative charge against it. Particles 1 and 2 deflect towards the negative plate, so they are positively charged. Particle 3 deflects towards the positive plate, so it is negatively charged.

    Charge-to-mass ratio. All three particles traverse the same field region, so each experiences a force $qE$ for a comparable time. The transverse deflection is proportional to the acceleration $a=qE/m$, and hence to the ratio $q/m$.

    Particle 3 shows the largest deflection over the same horizontal distance, so it undergoes the greatest acceleration and therefore has the highest charge-to-mass ratio.

    ✦ Particles 1 and 2 are positive, particle 3 is negative; particle 3 has the highest charge to mass ratio

14. 143 marksNCERT Exercises, Chapter 1

    Consider a uniform electric field $\vec{E}=3\times 10^3\hat{i}$ N/C. (a) What is the flux of this field through a square of 10 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?

    Hint. Flux is E A cos(theta), where theta is the angle between the field and the area's normal — not the plane.

    Electric flux through a flat surface is $\varphi =\vec{E}\cdot \vec{A}=EA\cos \theta$, where $\theta$ is measured between the field and the normal to the surface.

    The area is $A=(0.10{)}^2=10^{-2}$ m${}^2$.

    (a) A plane parallel to the $yz$ plane has its normal along the $x$-axis, which is the field direction, so $\theta =0$ and $\cos \theta =1$:

    $\varphi =3\times 10^3\times 10^{-2}=30{\text{N m}}^2/\text{C}$

    (b) Now $\theta =60^{\circ }$, so $\cos \theta =0.5$:

    $\varphi =3\times 10^3\times 10^{-2}\times 0.5=15{\text{N m}}^2/\text{C}$

    Using the angle the plane makes with the field, rather than the angle its normal makes, is the standard error here and would give $\sin$ instead of $\cos$.

    The cosine appears because only the component of the field along the normal carries flux through the surface.

    ✦ (a) 30 N m^2/C (b) 15 N m^2/C

15. 152 marksNCERT Exercises, Chapter 1

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

    Hint. A cube is a closed surface. Ask what charge it encloses.

    The cube is a closed surface, so Gauss's law applies: ${\varphi }_{\text{net}}=q_{\text{enclosed}}/{\epsilon }_0$.

    The field given is uniform, which means it is not produced by any charge inside the cube, so $q_{\text{enclosed}}=0$ and the net flux is zero.

    The same conclusion follows geometrically. Since the field points along $\hat{i}$, only the two faces perpendicular to the $x$-axis have any flux through them. Field lines enter one of those faces and leave the opposite one, and because the field is uniform the two contributions are equal in magnitude and opposite in sign, cancelling exactly.

    The remaining four faces have their normals perpendicular to the field, so they contribute nothing.

    Note that the side length of 20 cm never enters the calculation.

    ✦ Net flux = 0

16. 163 marksNCERT Exercises, Chapter 1

    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 $8.0\times 10^3$ Nm${}^2$/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?

    Hint. Gauss's law relates flux to the NET enclosed charge, which is the key to part (b).

    Gauss's law states $\varphi =q_{\text{enclosed}}/{\epsilon }_0$, so the enclosed charge follows immediately.

    (a) $q={\epsilon }_0\varphi =8.854\times 10^{-12}\times 8.0\times 10^3=7.08\times 10^{-8}$ C, that is about $0.07\mu$C.

    (b) No. Zero net flux tells us only that the net charge enclosed is zero, not that there is no charge at all.

    The box could contain equal amounts of positive and negative charge — say $+5\mu$C and $-5\mu$C — which sum to zero and therefore produce no net flux, while plenty of charge is present inside.

    Gauss's law is sensitive only to the algebraic sum of the enclosed charges.

    ✦ (a) q = 7.08 x 10^-8 C, about 0.07 microcoulomb (b) No — zero flux means zero NET charge, and equal positive and negative charges could both be present

17. 173 marksNCERT Exercises, Chapter 1

    A point charge $+10\mu$C is a distance 5 cm directly above the centre of a square of side 10 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 cm.)

    Hint. Follow the hint: the charge sits exactly at the centre of a cube whose face is this square.

    Follow the textbook's hint, which converts an awkward integral into a symmetry argument.

    A cube of edge 10 cm has its centre 5 cm from each face. Since the charge is 5 cm directly above the centre of the square, the charge sits exactly at the centre of such a cube, with the given square as one of its six faces.

    By Gauss's law the total flux through the whole closed cube is:

    ${\varphi }_{\text{total}}=\frac{q}{{\epsilon }_0}$

    By symmetry the six faces are equivalent, so each carries one sixth of the total:

    ${\varphi }_{\text{face}}=\frac{q}{6{\epsilon }_0}=\frac{10\times 10^{-6}}{6\times 8.854\times 10^{-12}}\approx 1.88\times 10^5{\text{N m}}^2/\text{C}$

    ✦ Flux through the square = 1.88 x 10^5 N m^2/C

18. 182 marksNCERT Exercises, Chapter 1

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

    Hint. Gauss's law depends only on the enclosed charge, not on the size or shape of the surface.

    Gauss's law gives the net flux through any closed surface as $\varphi =q_{\text{enclosed}}/{\epsilon }_0$.

    $\varphi =\frac{2.0\times 10^{-6}}{8.854\times 10^{-12}}\approx 2.26\times 10^5{\text{N m}}^2/\text{C}$

    The edge length of 9.0 cm is deliberately irrelevant, because the flux depends only on the charge enclosed and not at all on the size or shape of the Gaussian surface.

    A cube of edge 1 cm or 1 m centred on the same charge would give exactly the same answer, since every field line leaving the charge must cross whichever closed surface surrounds it.

    ✦ Net flux = 2.26 x 10^5 N m^2/C, independent of the cube's size

19. 193 marksNCERT Exercises, Chapter 1

    A point charge causes an electric flux of $-1.0\times 10^3$ Nm${}^2$/C to pass through a spherical Gaussian surface of 10.0 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?

    Hint. Part (a) tests the same independence as the previous question; part (b) inverts Gauss's law.

    (a) The flux through a closed surface depends only on the charge enclosed, not on the radius. Doubling the radius encloses exactly the same charge, so the flux is unchanged at $-1.0\times 10^3$ N m${}^2$/C.

    Physically, the field falls off as $1/r^2$ while the area grows as $r^2$, and the two effects cancel exactly.

    (b) Inverting Gauss's law:

    $q={\epsilon }_0\varphi =8.854\times 10^{-12}\times (-1.0\times 10^3)=-8.85\times 10^{-9}\text{C}$

    The negative sign of the flux is what tells us the charge is negative — field lines point inward, entering the surface rather than leaving it.

    ✦ (a) Unchanged, -1.0 x 10^3 N m^2/C (b) q = -8.85 x 10^-9 C, about -8.9 nC

20. 203 marksNCERT Exercises, Chapter 1

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

    Hint. Outside a uniformly charged sphere the field is that of a point charge at the centre. The inward direction fixes the sign.

    Outside a charged conducting sphere the field is identical to that of a point charge of the same total charge placed at the centre. The point of interest is at 20 cm, outside the 10 cm sphere, so this applies.

    $E=\frac{kq}{r^2}\Rightarrow q=\frac{E{r}^2}{k}=\frac{1.5\times 10^3\times (0.20{)}^2}{9\times 10^9}=\frac{60}{9\times 10^9}$

    which gives $|q|=6.67\times 10^{-9}$ C.

    The field points radially inward, and field lines run into negative charges. Therefore the charge is negative:

    $q=-6.67\times 10^{-9}\text{C}\approx -6.7\text{nC}$

    The sphere's own radius of 10 cm is not needed, since the field was measured outside it.

    ✦ q = -6.67 x 10^-9 C, about -6.7 nC

21. 213 marksNCERT Exercises, Chapter 1

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

    Hint. Halve the diameter to get the radius before computing the surface area.

    The diameter is given, so halve it first: $r=1.2$ m. Using the diameter directly is the usual slip here.

    (a) The surface area of the sphere is:

    $A=4\pi r^2=4\pi (1.2{)}^2\approx 18.1{\text{m}}^2$

    The charge is the surface charge density times the area:

    $q=\sigma A=80.0\times 10^{-6}\times 18.1\approx 1.45\times 10^{-3}\text{C}$

    (b) By Gauss's law the total flux leaving the surface is:

    $\varphi =\frac{q}{{\epsilon }_0}=\frac{1.45\times 10^{-3}}{8.854\times 10^{-12}}\approx 1.6\times 10^8{\text{N m}}^2/\text{C}$

    ✦ (a) q = 1.45 x 10^-3 C (b) flux = 1.6 x 10^8 N m^2/C

22. 223 marksNCERT Exercises, Chapter 1

    An infinite line charge produces a field of $9\times 10^4$ N/C at a distance of 2 cm. Calculate the linear charge density.

    Hint. Use the Gauss's law result for an infinite line charge, where E falls off as 1/r rather than 1/r squared.

    For an infinite line charge, applying Gauss's law with a cylindrical surface gives:

    $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$

    Note that the field falls off as $1/r$, not $1/r^2$ as it does for a point charge — using the wrong dependence is the main trap.

    Rearranging for the linear charge density:

    $\lambda =2\pi {\epsilon }_{0}rE=\frac{rE}{2k}=\frac{0.02\times 9\times 10^4}{2\times 9\times 10^9}=\frac{1800}{1.8\times 10^{10}}$

    so $\lambda =1.0\times 10^{-7}$ C/m, that is $0.1\mu$C/m.

    ✦ lambda = 1.0 x 10^-7 C/m = 0.1 microcoulomb per metre

23. 234 marksNCERT Exercises, Chapter 1

    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 $17.0\times 10^{-22}$ 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?

    Hint. Superpose the fields of the two sheets in each of the three regions, watching their directions.

    Treat each plate as an infinite charged sheet producing a field of magnitude $\sigma /2{\epsilon }_0$ on both sides, then superpose.

    The field of a positive sheet points away from it; that of a negative sheet points towards it.

    (a) Outside the first plate. The field from the positive sheet points one way and that from the negative sheet points the opposite way, with equal magnitudes $\sigma /2{\epsilon }_0$. They cancel exactly, so $E=0$.

    (b) Outside the second plate. By the identical argument the two contributions again cancel, so $E=0$.

    (c) Between the plates. Here both contributions point the same way, from the positive plate towards the negative plate, so they add:

    $E=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}=\frac{17.0\times 10^{-22}}{8.854\times 10^{-12}}\approx 1.92\times 10^{-10}\text{N/C}$

    directed from the positively charged plate to the negatively charged one. This is why the field of a parallel-plate capacitor is confined to the region between its plates.

    ✦ (a) E = 0 (b) E = 0 (c) E = 1.92 x 10^-10 N/C, directed from the positive plate to the negative plate

Solutions written by the tuition.in editorial team and checked against leph101.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.