NCERT Solutions · Class 12 Physics Electric Charges and Fields

By Coulomb's law, the force between two point charges in air is \[F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} = k\,\frac{q_1 q_2}{r^2},\qquad k = 9\times10^{9}\ \mathrm{N\,m^2\,C^{-2}}\] Substituting \(\displaystyle q_1 = 2\times10^{-7}\,\mathrm{C}\), \(\displaystyle q_2 = 3\times10^{-7}\,\mathrm{C}\) and \(\displaystyle r = 30\,\mathrm{cm} = 0.30\,\mathrm{m}\): \[F = \frac{(9\times10^{9})(2\times10^{-7})(3\times10^{-7})}{(0.30)^2} = \frac{5.4\times10^{-4}}{0.09}\] \[F = 6\times10^{-3}\,\mathrm{N}\] Both charges are positive, so the force is repulsive.

(a) Rearranging Coulomb's law for the separation: \[F = k\,\frac{q_1 q_2}{r^2}\ \Longrightarrow\ r^2 = k\,\frac{q_1 q_2}{F}\] Taking magnitudes, with \(\displaystyle q_1 = 0.4\,\mu\mathrm{C} = 4\times10^{-7}\,\mathrm{C}\), \(\displaystyle q_2 = 0.8\,\mu\mathrm{C} = 8\times10^{-7}\,\mathrm{C}\) and \(\displaystyle F = 0.2\,\mathrm{N}\): \[r^2 = \frac{(9\times10^{9})(4\times10^{-7})(8\times10^{-7})}{0.2} = \frac{2.88\times10^{-3}}{0.2} = 1.44\times10^{-2}\,\mathrm{m^2}\] \[r = 0.12\,\mathrm{m} = 12\,\mathrm{cm}\](b) By Newton's third law the two spheres exert equal and opposite forces on each other, so the force on the second sphere due to the first is also \[F = 0.2\,\mathrm{N}\] The charges are of opposite sign, so it is attractive.

Dimensions. Coulomb's law gives \(\displaystyle k\,e^2 = F r^2\), and Newton's law of gravitation gives \(\displaystyle G\,m_e m_p = F r^2\). Both numerator and denominator therefore have dimensions of \(\displaystyle \mathrm{N\,m^2}\), i.e. \(\displaystyle \mathrm{[M\,L^3\,T^{-2}]}\), so the ratio is dimensionless.Value. Using \(\displaystyle k = 9\times10^{9}\,\mathrm{N\,m^2\,C^{-2}}\), \(\displaystyle e = 1.6\times10^{-19}\,\mathrm{C}\), \(\displaystyle G = 6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\), \(\displaystyle m_e = 9.11\times10^{-31}\,\mathrm{kg}\), \(\displaystyle m_p = 1.67\times10^{-27}\,\mathrm{kg}\): \[k e^2 = (9\times10^{9})(1.6\times10^{-19})^2 = 2.30\times10^{-28}\] \[G m_e m_p = (6.67\times10^{-11})(9.11\times10^{-31})(1.67\times10^{-27}) = 1.01\times10^{-67}\] \[\frac{k e^2}{G m_e m_p} = \frac{2.30\times10^{-28}}{1.01\times10^{-67}} \approx 2.3\times10^{39}\]Meaning. This is the ratio of the electrostatic force to the gravitational force between an electron and a proton, at any separation — both fall off as \(\displaystyle 1/r^2\), so \(\displaystyle r\) cancels. The electrical force is about \(\displaystyle 10^{39}\) times stronger, which is why gravity is completely negligible in atomic structure.

(a) Quantisation of charge means the charge on any body is not arbitrary: it is always an integral multiple of the elementary charge \(\displaystyle e\),\[q = n e,\qquad n = 0,\ \pm 1,\ \pm 2,\ \ldots,\qquad e = 1.6\times10^{-19}\,\mathrm{C}\] Charge is transferred only by whole electrons, so no body can carry a charge of, say, \(\displaystyle 1.5\,e\).(b) Because \(\displaystyle e\) is minute compared with the charges met in practice. A charge of \(\displaystyle 1\,\mu\mathrm{C}\) corresponds to \[n = \frac{10^{-6}}{1.6\times10^{-19}} \approx 6\times10^{12}\] elementary charges. Adding or removing one electron changes such a charge by about one part in \(\displaystyle 10^{13}\) — far below anything measurable — so at the macroscopic scale charge behaves as though it varies continuously, and quantisation can be ignored.

Rubbing does not create charge; it only transfers electrons from one body to the other. Glass and silk are electrically neutral to begin with, each containing equal amounts of positive and negative charge.When they are rubbed together, some electrons are transferred from the glass to the silk. The glass, having lost electrons, is left with a net positive charge \(\displaystyle +q\); the silk, having gained exactly those same electrons, acquires \(\displaystyle -q\). The two charges are equal in magnitude and opposite in sign, so the total charge of the system is \[(+q) + (-q) = 0\] which is what it was before rubbing. The same argument applies to every such pair of bodies.So the appearance of charge on both bodies is not a creation of charge but a redistribution of it, and the observation is exactly what the law of conservation of charge requires.

Take the square ABCD of side \(\displaystyle 10\,\mathrm{cm}\) with the test charge \(\displaystyle q = 1\,\mu\mathrm{C}\) at its centre O. Every corner is the same distance from O, namely half the diagonal: \[r = \frac{1}{\sqrt{2}}\times 0.10\,\mathrm{m}\]A and C are opposite corners and carry the same charge, \(\displaystyle q_A = q_C = +2\,\mu\mathrm{C}\). They are equidistant from O and lie on opposite sides of it, so the two forces they exert on the charge at O are equal in magnitude and opposite in direction, and they cancel.The same is true of B and D, which both carry \(\displaystyle -5\,\mu\mathrm{C}\): equal magnitudes, opposite directions, so they cancel too.Adding the four contributions: \[\vec{F} = (\vec{F}_A + \vec{F}_C) + (\vec{F}_B + \vec{F}_D) = \vec{0} + \vec{0}\] \[F = 0\,\mathrm{N}\]The force on the \(\displaystyle 1\,\mu\mathrm{C}\) charge at the centre is zero. Note this follows from symmetry alone — the magnitudes of the charges never had to be used.

(a) A field line is drawn so that its tangent at every point gives the direction of \(\displaystyle \vec{E}\) there, and it traces the path a free positive test charge would begin to follow. The electrostatic field has a definite magnitude and direction at every point of space (except exactly at a point charge), so the direction the line must take is defined everywhere along it.A sudden break would mean the field abruptly ceased to exist at that point and reappeared elsewhere, which would require the test charge to jump discontinuously. Since \(\displaystyle \vec{E}\) is continuous in charge-free space, the line cannot break.(b) If two field lines crossed at a point, then at that single point one could draw two different tangents, and hence the field would have two different directions there simultaneously. A test charge placed at that point would have to move in two directions at once, which is impossible.The electric field at any point has one unique direction, so two field lines can never intersect.

Let O be the midpoint of AB, so \(\displaystyle r = \dfrac{20\,\mathrm{cm}}{2} = 0.10\,\mathrm{m}\) from each charge.(a) The field of the positive charge \(\displaystyle q_A = 3\,\mu\mathrm{C}\) at O points away from A, i.e. from A towards B. Its magnitude is \[E_A = \frac{k q_A}{r^2} = \frac{(9\times10^{9})(3\times10^{-6})}{(0.10)^2} = 2.7\times10^{6}\,\mathrm{N\,C^{-1}}\] The field of the negative charge \(\displaystyle q_B = -3\,\mu\mathrm{C}\) at O points towards B — that is, in the same direction, from A towards B — with the same magnitude \[E_B = 2.7\times10^{6}\,\mathrm{N\,C^{-1}}\] The two contributions therefore add rather than cancel: \[E = E_A + E_B = 5.4\times10^{6}\,\mathrm{N\,C^{-1}}\] directed along AB, from A towards B (i.e. along OB).(b) The force on a test charge of magnitude \(\displaystyle q = 1.5\times10^{-9}\,\mathrm{C}\) is \[F = qE = (1.5\times10^{-9})(5.4\times10^{6}) = 8.1\times10^{-3}\,\mathrm{N}\] The test charge is negative, so the force is opposite to \(\displaystyle \vec{E}\): it acts along BA, i.e. from O towards A.

Total charge. \[q_{\text{total}} = q_A + q_B = (2.5\times10^{-7}) + (-2.5\times10^{-7}) = 0\] The system is neutral overall — it is a pure dipole.Dipole moment. The two charges are equal and opposite, separated by \[2a = 15\,\mathrm{cm} + 15\,\mathrm{cm} = 30\,\mathrm{cm} = 0.30\,\mathrm{m}\] so \[p = q\,(2a) = (2.5\times10^{-7})(0.30) = 7.5\times10^{-8}\,\mathrm{C\,m}\]Direction. The dipole moment points from the negative charge to the positive charge. Here the negative charge sits at \(\displaystyle z = +15\,\mathrm{cm}\) and the positive at \(\displaystyle z = -15\,\mathrm{cm}\), so \(\displaystyle \vec{p}\) points along the negative \(\displaystyle z\)-direction: \[\vec{p} = 7.5\times10^{-8}\,\hat{(-z)}\ \mathrm{C\,m}\]

The torque on a dipole in a uniform field is \[\vec{\tau} = \vec{p}\times\vec{E},\qquad \tau = pE\sin\theta\] Substituting \(\displaystyle p = 4\times10^{-9}\,\mathrm{C\,m}\), \(\displaystyle E = 5\times10^{4}\,\mathrm{N\,C^{-1}}\) and \(\displaystyle \theta = 30^\circ\): \[\tau = (4\times10^{-9})(5\times10^{4})\sin 30^\circ = (2\times10^{-4})(0.5)\] \[\tau = 1\times10^{-4}\,\mathrm{N\,m}\]The torque acts to rotate the dipole into alignment with the field.