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Rajasthan Board Class 12 (RBSE) Electric Charges and Fields — practice questions

Explore 4 free Electric Charges and Fields questions in Rajasthan Board Class 12 (RBSE) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Electric Charges and Fields questions and exercises with answers

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Worked practice questions

The smallest charge that can exist freely in nature is

  1. $+1\,\text{C}$
  2. $e = 1.6\times 10^{-19}\,\text{C}$
  3. the mass of an electron
  4. zero
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Correct answer: B. $e = 1.6\times 10^{-19}\,\text{C}$

1. syllabus states the QUANTISATION of electric charge: $q = ne$ for some integer $n$. 2. The elementary charge is $e = 1.6\times 10^{-19}\,\text{C}$ — the magnitude of the charge on an electron (or a proton). 3. Free quarks would carry fractional charge, but they are confined inside hadrons and never observed in isolation. So $e$ is the smallest FREE charge. 4. Macroscopic charges always come as integer multiples of $e$. 5. Option A is a huge macroscopic charge. Option C is a MASS, not a charge. Option D contradicts the existence of charged objects. _Source: syllabus Physics Part 1, Ch 1 "Electric Charges and Fields", §1.4 (Quantisation of Charge) + §1.5, p. 4–5._

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An electric dipole of dipole moment $\vec{p}$ is placed in a UNIFORM electric field $\vec{E}$. The torque experienced is

  1. $\vec{p} + \vec{E}$
  2. $\vec{p} \cdot \vec{E}$
  3. $\vec{p} \times \vec{E}$
  4. $|\vec{p}|\,|\vec{E}|\,\hat{r}$
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Correct answer: C. $\vec{p} \times \vec{E}$

1. syllabus derives the torque on a dipole in a uniform field. 2. The two equal-and-opposite forces $\pm qE$ at the dipole ends form a COUPLE. Net force is zero, but the couple produces a torque. 3. The torque magnitude is $\tau = pE\sin\theta$, where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. 4. Direction: along $\vec{p}\times\vec{E}$ by the right-hand rule — perpendicular to the plane containing $\vec{p}$ and $\vec{E}$. 5. So vectorially $\vec{\tau} = \vec{p}\times\vec{E}$. The torque acts to ROTATE the dipole until $\vec{p} \parallel \vec{E}$ (stable equilibrium). 6. Option A has wrong dimensions. Option B is the scalar product (a number, not a vector). Option D is dimensionally wrong. _Source: syllabus Physics Part 1, Ch 1, §1.10 (Dipole in a Uniform Field), p. 25._

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The electric field at a distance $r$ from a POINT charge $q$ (in vacuum) has magnitude

  1. $\dfrac{kq}{r^2}$
  2. $\dfrac{kq}{r}$
  3. $\dfrac{kq^2}{r^2}$
  4. $kq\,r^2$
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Correct answer: A. $\dfrac{kq}{r^2}$

1. syllabus defines the electric field due to a point charge: $\vec{E} = \dfrac{kq}{r^2}\,\hat{r}$, where $k = \dfrac{1}{4\pi\varepsilon_0}$. 2. Inverse-SQUARE in distance — same form as the gravitational field of a point mass. 3. Linear in the source charge $q$. Direction: radially outward if $q > 0$, inward if $q < 0$. 4. Option B gives the form of the POTENTIAL ($\sim 1/r$), not the field. Option C is wrong — only one $q$ appears in $E$ (the source charge) a second charge enters in the FORCE expression. Option D has the wrong distance dependence. _Source: syllabus Physics Part 1, Ch 1, §1.7 (Electric Field), p. 11–12._

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