The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N. (a) What is the distance between them? (b) What is the force on the second sphere due to the first?
(b) By Newton's third law, the force on sphere 2 is also 0.2 N, directed toward sphere 1 (attractive).
Exercise 1.3
Check whether the ratio \(ke^2/Gm_e m_p\) is dimensionless and compute its numerical value for an electron–proton pair. What does it signify?
\([ke^2] = \) N·m²/C² × C² = N·m²; \([Gm_em_p] = \) N·m²/kg² × kg² = N·m². Ratio is dimensionless.
\[\frac{ke^2}{Gm_em_p} = \frac{(9\times 10^9)(1.6\times 10^{-19})^2}{(6.67\times 10^{-11})(9.1\times 10^{-31})(1.67\times 10^{-27})}\approx \boxed{2.3\times 10^{39}}\]
Significance: at the atomic level, Coulomb force is about \(10^{39}\) times stronger than gravity.
Exercise 1.4
(a) Explain the meaning of "quantisation of charge". (b) Why can quantisation of charge be ignored when dealing with macroscopic charges (like 1 C)?
(a) Any free charge is an integer multiple of the elementary charge: \(q = ne\), \(e = 1.6\times 10^{-19}\) C.
(b) \(n = 1\,\text{C}/(1.6\times 10^{-19}) = 6.25\times 10^{18}\). One step \(e\) compared to 1 C is \(1.6\times 10^{-19}\) — utterly imperceptible. At this scale, the discrete staircase looks smooth/continuous.
Exercise 1.5
When a glass rod is rubbed with silk, both acquire charges of equal magnitude but opposite sign. Is this consistent with the law of conservation of charge?
Yes. Before rubbing, both were neutral (net charge zero). After rubbing, some electrons transferred from glass to silk — silk becomes negative, glass becomes positive by the same amount. Total charge of the isolated (glass + silk) system remains zero. No charge is created or destroyed; only redistributed. This is exactly conservation of charge.
Exercise 1.6
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?
Diagonally-opposite charges A and C are identical (+2 μC) — their forces on the central charge are equal and opposite and cancel. Similarly, B and D (–5 μC) cancel.
\[\boxed{\vec F_\text{net} = 0}\]
Exercise 1.7
(a) An electrostatic field line is a continuous curve — why? Can a field line have sudden breaks? (b) Explain why two field lines never cross each other.
(a) A charged particle placed in an electric field experiences a continuous force and hence moves along a continuous path. A field line has no break because the electric field exists continuously at every point in space (except at the location of a point charge); a break would mean no field at that point, which is physically impossible.
(b) If two lines intersected, the electric field at the intersection point would have two different directions (one tangent per line). A field vector can have only one direction at a point; therefore, intersecting field lines are forbidden.
Exercise 1.8
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 them? (b) If a negative test charge of magnitude \(1.5\times 10^{-9}\) C is placed at O, what force is experienced by it?
Midpoint O is 10 cm from each charge. Each field points from the positive toward the negative (they add):
\[E = 2\times\frac{kq}{r^2} = 2\times \frac{(9\times 10^9)(3\times 10^{-6})}{(0.10)^2} = \boxed{5.4\times 10^6\,\text{N/C}}\]
Direction: from A (+) toward B (–).
Force on –1.5 nC test charge: \(F = qE = (1.5\times 10^{-9})(5.4\times 10^6) = \boxed{8.1\times 10^{-3}\,\text{N}}\), directed from B toward A (opposite to E, because charge is negative).
Exercise 1.9
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 is the total charge and the electric dipole moment of the system?
Total charge: \(+2.5\times 10^{-7} + (-2.5\times 10^{-7}) = \boxed{0}\).
Separation between A and B: 30 cm = 0.30 m. Dipole moment points from –q (at B) to +q (at A), i.e. along –z:
\[p = q\cdot 2a = (2.5\times 10^{-7})(0.30) = \boxed{7.5\times 10^{-8}\,\text{C·m along }-\hat z}\]
Exercise 1.10
An electric dipole with dipole moment \(4\times 10^{-9}\) C·m is aligned at 30° with the direction of a uniform external electric field of magnitude \(5\times 10^4\) N/C. Calculate the magnitude of the torque acting on the dipole.
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?
(a) \(n = |q|/e = 3\times 10^{-7}/(1.6\times 10^{-19}) = \boxed{1.875\times 10^{12}\ \text{electrons}}\). Electrons are transferred from wool to polythene (polythene becomes negative).
(b) Yes, but negligibly small. Mass transferred \(= n\times m_e = 1.875\times 10^{12}\times 9.1\times 10^{-31} = \boxed{1.7\times 10^{-18}\,\text{kg}}\) — too tiny to measure.
Exercise 1.12
(a) Two insulated charged copper spheres A and B have identical sizes and charges \(6.5\times 10^{-7}\) C each. The distance between their centres is 50 cm. Find the force of repulsion. (b) If the mass of each sphere is 0.1 g, determine the gravitational acceleration \(g\) in the problem if the electric force equals the weight of one sphere when they are placed 50 cm apart.
(b) If this electric repulsion equals the weight \(mg\): \(g = F/m = 1.52\times 10^{-2}/(0.1\times 10^{-3}) = \boxed{152\,\text{m/s}^2}\).
(The original NCERT question asks for the charge given F and d, then if mass = 0.1 g, calculate g — interpretation shown.)
Exercise 1.13
Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Each sphere originally carries charge \(Q = 6.5\times 10^{-7}\) C.
Step 1: uncharged sphere touches A → they share equally → A and the third each get \(Q/2\).
Step 2: third (\(Q/2\)) touches B (\(Q\)) → they share → each gets \((Q/2 + Q)/2 = 3Q/4\).
Final charges: A has \(Q/2\), B has \(3Q/4\).
\[F' = \frac{k(Q/2)(3Q/4)}{r^2} = \frac{3}{8}\cdot\frac{kQ^2}{r^2} = \frac{3}{8}F_\text{original}\]
\[F' = (3/8)\times 1.52\times 10^{-2} = \boxed{5.7\times 10^{-3}\,\text{N}}\]
Exercise 1.14
Figure 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?
If the field points downward in the figure, particles that deflect upward carry negative charge (force opposite to E) and those deflecting downward are positive.
Typically: particles 1 and 2 deflect one way (e.g. up) and particle 3 the other (down). In the standard NCERT picture, particles 1 and 2 are negative, particle 3 is positive.
The particle that deflects most (sharpest curvature) has the largest acceleration \(a = qE/m\), hence the largest \(q/m\). In NCERT Fig. 1.34 that is particle 3.
Exercise 1.15
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 side 10 cm whose plane is parallel to the y–z plane? (b) What is the flux through the same square if the normal to its plane makes an angle of 60° with the x-axis?
\(A = 0.10\times 0.10 = 0.01\) m².
(a) Plane parallel to y–z → normal along x-axis, parallel to \(\vec E\). \(\theta = 0°\):
\(\Phi = EA\cos 0° = 3\times 10^3\times 0.01 = \boxed{30\,\text{N·m}^2/\text{C}}\).
What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
For a uniform field, flux entering the cube equals flux leaving. Net enclosed charge is zero.
\[\boxed{\Phi_\text{net} = 0}\]
Explicitly: the two faces perpendicular to \(\vec E\) have fluxes \(+EA\) and \(-EA\) which cancel; the other four have \(\Phi = 0\).
Exercise 1.17
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface is \(8.0\times 10^3\) N·m²/C. (a) What is the net charge inside the box? (b) If the net outward flux were zero, could you conclude that there were no charges inside the box?
(b) No. Zero flux only means net charge is zero — there could be equal amounts of positive and negative charge inside.
Exercise 1.18
A point charge of \(+10\,\mu\)C is placed at the centre of a cube of side 10 cm. Find the electric flux through (a) the whole cube, (b) one face of the cube.
(b) By symmetry, each of 6 faces receives the same flux:
\(\Phi_\text{face} = \Phi_\text{total}/6 = \boxed{1.88\times 10^5\,\text{N·m}^2/\text{C}}\).
Exercise 1.19
A point charge causes an electric flux of \(-1.0\times 10^3\) N·m²/C to pass through a spherical Gaussian surface of 10 cm radius centred on the charge. (a) If the radius were doubled, what would be the flux? (b) What is the value of the point charge?
(a) Flux depends only on enclosed charge, not radius. So if radius is doubled, \(\Phi\) is unchanged at \(-1.0\times 10^3\) N·m²/C.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field at a distance 20 cm from its centre is \(1.5\times 10^3\) N/C directed radially inward, what is the net charge on the sphere?
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m². (a) Find the charge on the sphere. (b) Find the total electric flux leaving the surface.
Radius \(R = 1.2\) m. \(A = 4\pi R^2 = 4\pi (1.2)^2 = 18.09\) m².
(a) \(Q = \sigma A = (80\times 10^{-6})(18.09) = \boxed{1.45\times 10^{-3}\,\text{C}}\).
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.
\[\lambda = 2\pi\varepsilon_0 r E = \frac{r E}{2k} = \frac{(0.02)(9\times 10^4)}{2\times 9\times 10^9} = \boxed{10^{-7}\,\text{C/m} = 0.1\,\mu\text{C/m}}\]
Exercise 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 \(17.0\times 10^{-22}\) C/m². What is E (a) in the outer region of the first plate, (b) in the outer region of the second plate, (c) between the plates?
(a) and (b) Outside the pair — the two plate fields are equal and opposite, so they cancel: \(\boxed{E = 0}\).
(c) Between plates — the two fields add:
\[E = \frac{\sigma}{\varepsilon_0} = \frac{17\times 10^{-22}}{8.854\times 10^{-12}} = \boxed{1.92\times 10^{-10}\,\text{N/C}}\]
directed from the positive plate to the negative plate.
Exercise 1.24
An oil drop of 12 excess electrons is held stationary under a constant electric field of \(2.55\times 10^4\) N/C (Millikan experiment). The density of the oil is 1.26 g/cm³. Estimate the radius of the drop (g = 9.81 m/s²).
Balance: electric force = weight → \(qE = mg = \rho V g = \rho \cdot \tfrac{4}{3}\pi r^3 g\).
\(q = 12e = 12\times 1.6\times 10^{-19} = 1.92\times 10^{-18}\) C.
\[r^3 = \frac{3qE}{4\pi\rho g} = \frac{3(1.92\times 10^{-18})(2.55\times 10^4)}{4\pi(1260)(9.81)}
= \frac{1.47\times 10^{-13}}{1.553\times 10^5} = 9.46\times 10^{-19}\,\text{m}^3\]
\[r = (9.46\times 10^{-19})^{1/3} \approx \boxed{9.8\times 10^{-7}\,\text{m} = 0.98\,\mu\text{m}}\]
Frequently Asked Questions - NCERT Exercises and Solutions: Electric Charges and Fields
What are the key NCERT exercise types in Chapter 1 Electric Charges and Fields?
NCERT Class 12 Physics Chapter 1 Electric Charges and Fields exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Electric Charges and Fields?
For numerical problems in NCERT Class 12 Physics Chapter 1 Electric Charges and Fields: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 1?
From NCERT Class 12 Physics Chapter 1 (Electric Charges and Fields), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 1 Electric Charges and Fields problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 1 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 1 Electric Charges and Fields exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 1 Electric Charges and Fields solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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