Match List-I with List-II.
| List-I | List-II |
|---|---|
| (A) The field at a point where a charge of \(2\ \mu\mathrm{C}\) feels a force of \(6\times10^{-3}\ \mathrm{N}\) | (I) \(2\times10^{4}\ \mathrm{N\,C^{-1}}\) |
| (B) The field at a distance of 3 cm from a point charge of 1 nC | (II) \(3\times10^{3}\ \mathrm{N\,C^{-1}}\) |
| (C) The field at the mid-point between charges of \(+2\) nC and \(-2\) nC placed 6 cm apart | (III) \(4\times10^{4}\ \mathrm{N\,C^{-1}}\) |
| (D) The vertical field that holds at rest a particle of mass 2 mg carrying a charge of 1 nC | (IV) \(1\times10^{4}\ \mathrm{N\,C^{-1}}\) |
(Take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\) and \(g=10\ \mathrm{m\,s^{-2}}\).)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
The electric field at a point is the force per unit positive charge there, \( E=F/q \). For a point charge it is \( E=Q/4\pi\varepsilon_0r^{2} \), and the fields of several charges add as vectors.
(A) The field where \( 2\ \mu\mathrm{C} \) feels \( 6\times10^{-3}\ \mathrm{N} \) — This matches (II).
\( \begin{aligned} E &= \frac{F}{q} \\ &= \frac{6\times10^{-3}}{2\times10^{-6}} \\ &= 3\times10^{3}\ \mathrm{N\,C^{-1}} \end{aligned} \)
(B) The field 3 cm from a point charge of 1 nC — This matches (IV).
\( \begin{aligned} E &= \frac{Q}{4\pi\varepsilon_0r^{2}} \\ &= \frac{9\times10^{9}\times1\times10^{-9}}{(0.03)^{2}} \\ &= 1\times10^{4}\ \mathrm{N\,C^{-1}} \end{aligned} \)
(C) The field midway between \( +2 \) nC and \( -2 \) nC, 6 cm apart — This matches (III). Each charge is 3 cm from the mid-point. Both fields point from the positive charge towards the negative charge, so they add.
\( \begin{aligned} E &= 2\times\frac{9\times10^{9}\times2\times10^{-9}}{(0.03)^{2}} \\ &= 2\times2\times10^{4} \\ &= 4\times10^{4}\ \mathrm{N\,C^{-1}} \end{aligned} \)
(D) The field that holds a particle of mass 2 mg and charge 1 nC at rest — This matches (I). The electric force \( qE \) must balance the weight \( mg \), with \( m=2\times10^{-6}\ \mathrm{kg} \).
\( \begin{aligned} E &= \frac{mg}{q} \\ &= \frac{2\times10^{-6}\times10}{1\times10^{-9}} \\ &= 2\times10^{4}\ \mathrm{N\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Electric Field – “… The force F is then negligibly small but the ratio F/q is finite and defines the electric field:”
\( \displaystyle E=\lim_{q\to0}\frac{F}{q} \qquad (1.9) \)
Electric Field – “The electric field produced by the charge Q at a point r is given as”
\( \displaystyle E(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^{2}}\,\hat r \)\( \qquad (1.6) \)
Electric field due to a system of charges – “By the superposition principle, the electric field E at r due to the system of charges is … E(r) = E1 (r) + E2 (r) + … + En(r)”