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A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$) shown in the figure is:
(Take $\dfrac{1}{4\pi\epsilon_0} = 9\times 10^9$ SI units)
Inside/on a charged shell, $V$ is constant — $V_C = V_P$.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The potential ($V$) at any axial point, at 2 m distance($r$) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 \times 10^{-6}\ \text{C m}$, is $\pm 9\times 10^3\ V$.
(Take $\dfrac{1}{4\pi\epsilon_0} = 9\times 10^9$ SI units)
Reason R: $V = \pm \dfrac{2P}{4\pi\epsilon_0 r^2}$, where $r$ is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
Correct axial potential: $V = p/(4\pi\epsilon_0 r^2) = 9\times 10^3$ V (A correct). R's formula has extra factor 2 (that is for field, not potential) — R false.
In the following circuit, the equivalent capacitance between terminal $A$ and terminal $B$ is:
Balanced bridge → middle $2\,\mu F$ carries no charge. Two arms each give $1\,\mu F$ in series; in parallel: $2\,\mu F$.
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:
Battery keeps $V$ constant; $C, Q, QV$ all increase; $U = \tfrac12 CV^2$ increases too — so B (decrease) wrong; ratio $Q/V = C$ changes — D wrong.
The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If $F_A$ and $F_B$ are the forces applied by the breaks on cars A and B, respectively, then the ratio $F_A/F_B$ is:
Work–energy: $F = \dfrac{KE}{d}$. $F_A = \dfrac{100}{1000} = 0.1$, $F_B = \dfrac{225}{1500} = 0.15$; $\dfrac{F_A}{F_B} = \dfrac{0.1}{0.15} = \dfrac{2}{3}$.
The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\frac{3}{8}d$ and $\frac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25\,K_2$, the value of $K_1$ is:
Series of three regions (air gap $= d/8$): $\dfrac{d}{2} = \dfrac{3d}{8K_1}+\dfrac{d}{2K_2}+\dfrac{d}{8}$. With $K_2 = 0.8K_1$: $\dfrac{3}{8K_1}+\dfrac{0.625}{K_1} = \dfrac{3}{8}\Rightarrow K_1 = \dfrac{8}{3} \approx 2.66$.
Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is F. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres A and B (radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:
After touching A: $q_A = q/2$. Touching B (now $q/2 + q = 3q/2$ shared): $q_B = 3q/4$. New $F\propto\left(\tfrac{q}{2}\right)\left(\tfrac{3q}{4}\right) = \tfrac{3}{8}q^2$, i.e. $\dfrac{3F}{8}$.
An electric dipole with dipole moment $5\times10^{-6}$ Cm is aligned with the direction of a uniform electric field of magnitude $4\times10^5$ N/C. The dipole is then rotated through an angle of $60^\circ$ with respect to the electric field. The change in the potential energy of the dipole is:
$\Delta U = pE(1-\cos60^\circ) = (5\times10^{-6})(4\times10^5)(1-0.5) = 2\times0.5 = 1.0$ J.
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