Electrostatics MCQs for NEET — Physics Questions with Answers

(c) Net charge on one H-atom 

=-e+e+$∆e$

Net electrostatic repulsive force between two H-atoms

$F_e = \frac{K{\left(∆e\right)}^2}{d^2}$

$F_G=\frac{G{m}_1^2}{d^2}$

It is given that 

$F_e-F_G=0$

$\Rightarrow \frac{K{\left(∆e\right)}^2}{d^2}-\frac{G{m}_1^2}{d^2}=0$

$\Rightarrow {\left(∆e\right)}^2=\frac{\left(6.67\times {10}^{-11}\right){\left(1.67\times {10}^{-27}\right)}^2}{9\times {10}^9}$

$\Rightarrow ∆e=1.437\times {10}^{-37}C$

(b) $\because$Torque on an electric dipole in an electric field,

             $\tau$=p$\times E\Rightarrow \left|\tau \right|=pE \sin \theta$

where $\theta$ is the angle between E and p

$$\begin{gathered} \Rightarrow 4=\rho \times 2\times {10}^5\times \frac{1}{2} \\ \Rightarrow p=4\times {10}^{-5}cm \\ \Rightarrow p=q2l \\ \therefore q2l =4\times {10}^{-5} \end{gathered}$$

Where 2l=2cm=$2\times {10}^{-2} m$

$q=\frac{4\times {10}^{-5}}{2\times {10}^{-2}}$

$\Rightarrow 2\times {10}^{-3}C=2mC$

Charge enclosed=q/8

Therefore,flux $\varphi =\frac{q_{enclosed}}{{\epsilon }_0}$

                    $\varphi =\frac{q}{8{\epsilon }_0}$

When the parallel metal plates are dipped in kerosene oil, the electric field between the plates decreases. This is because the kerosene oil has a higher permittivity than vacuum or air, which reduces the effective electric field according to the relation E = E₀/κ, where κ is the relative permittivity.

Energy stored in capacitor = Heat produced in process of charging

Initial charge on the spheres:

$$\begin{gathered} {\mathrm{Q}}_1={\mathrm{C}}_1{\mathrm{V}}_1 \& {\mathrm{Q}}_2={\mathrm{C}}_2{\mathrm{V}}_2 \\ \mathrm{Initial} \mathrm{potential} \mathrm{energy} \mathrm{of} \mathrm{the} \mathrm{spheres}- \\ {\mathrm{U}}_1=\frac{1}{2}{\mathrm{C}}_1{\mathrm{V}}_1^2 and, {\mathrm{U}}_2=\frac{1}{2}{\mathrm{C}}_2{\mathrm{V}}_2^2 \\ \mathrm{Charges} \mathrm{will} \mathrm{flow} \mathrm{between} \mathrm{the} \mathrm{spheres} \mathrm{until} \mathrm{they} \mathrm{have} \mathrm{same} \mathrm{electric} \mathrm{potential}. \\ \mathrm{Let} \mathrm{q} \mathrm{charge} \mathrm{flows} \mathrm{from} \mathrm{sphere} 1 \mathrm{to} 2. \\ \mathrm{Then}, {\mathrm{V}}_1\text{'}={\mathrm{V}}_2\text{'} \\ \frac{{\mathrm{Q}}_1-\mathrm{q}}{{\mathrm{C}}_1}=\frac{{\mathrm{Q}}_2+\mathrm{q}}{{\mathrm{C}}_2} \\ \Rightarrow \mathrm{q}=\frac{{\mathrm{Q}}_1{\mathrm{C}}_2-{\mathrm{Q}}_2{\mathrm{C}}_1}{{\mathrm{C}}_1+{\mathrm{C}}_2} = \frac{{\mathrm{C}}_1{\mathrm{V}}_1{\mathrm{C}}_2-{\mathrm{C}}_2{\mathrm{V}}_2{\mathrm{C}}_1}{{\mathrm{C}}_1+{\mathrm{C}}_2} \\ \Rightarrow \mathrm{q}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}\left({\mathrm{V}}_1-{\mathrm{V}}_2\right) \\ \Rightarrow {\mathrm{V}}_1\text{'}={\mathrm{V}}_2\text{'}=\frac{1}{{\mathrm{C}}_1}\left[{\mathrm{Q}}_1-\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}\left({\mathrm{V}}_1-{\mathrm{V}}_2\right)\right] \\ =\frac{1}{{\mathrm{C}}_1}\left[\frac{{\mathrm{C}}_1{\mathrm{V}}_1\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)-{\mathrm{C}}_1{\mathrm{C}}_2\left({\mathrm{V}}_1-{\mathrm{V}}_2\right)}{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}\right] \\ =\frac{1}{{\mathrm{C}}_1}\left[\frac{{\mathrm{C}}_1^2{\mathrm{V}}_1-{\mathrm{C}}_1{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}\right] \\ =\frac{\left({\mathrm{C}}_1{\mathrm{V}}_1-{\mathrm{C}}_2{\mathrm{V}}_2\right)}{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)} \\ \mathrm{Final} \mathrm{Potential} \mathrm{enrergy} \mathrm{of} \mathrm{the} \mathrm{spheres}- \\ {\mathrm{U}}_1\text{'}=\frac{1}{2}{\mathrm{C}}_1{{\mathrm{V}}_1^{\text{'}}}^2=\frac{1}{2}{\mathrm{C}}_1{\left[\frac{{\mathrm{C}}_1{\mathrm{V}}_1-{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}\right]}^2 \\ {\mathrm{U}}_2\text{'}=\frac{1}{2}{\mathrm{C}}_2{{\mathrm{V}}_2^{\text{'}}}^2=\frac{1}{2}{\mathrm{C}}_2{\left[\frac{{\mathrm{C}}_1{\mathrm{V}}_1-{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}\right]}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Heat} \mathrm{produced} = \mathrm{Loss} \mathrm{in} \mathrm{potential} \mathrm{energy} \\ =\mathrm{Initial} \mathrm{potential} \mathrm{enrgy} - \mathrm{Final} \mathrm{potential} \mathrm{enrgy} = \left({\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}\right) \\ =\left[\left({\mathrm{U}}_1+{\mathrm{U}}_2\right)-\left({\mathrm{U}}_1\text{'}+{\mathrm{U}}_2\text{'}\right)\right] \\ =\left[\frac{1}{2}\left({\mathrm{C}}_1{\mathrm{V}}_1^2+{\mathrm{C}}_2{\mathrm{V}}_2^2\right)-\frac{1}{2}\left({\mathrm{C}}_1+{\mathrm{C}}_2\right){\left[\frac{{\mathrm{C}}_1{\mathrm{V}}_1-{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}\right]}^2\right] \\ =\frac{1}{2}\left[\frac{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)\left({\mathrm{C}}_1{\mathrm{V}}_1^2+{\mathrm{C}}_2{\mathrm{V}}_2^2\right)}{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}-{\left({\mathrm{C}}_1{\mathrm{V}}_1-{\mathrm{C}}_2{\mathrm{V}}_2\right)}^2\right] \\ =\frac{-1}{2}\left[\frac{{\mathrm{C}}_1^2{\mathrm{V}}_1^2+{\mathrm{C}}_2^2{\mathrm{V}}_2^2-2{\mathrm{C}}_1{\mathrm{V}}_1{\mathrm{C}}_2{\mathrm{V}}_2-{\mathrm{C}}_1^2{\mathrm{V}}_1^2-{\mathrm{C}}_1{\mathrm{C}}_2{\mathrm{V}}_2^2-{\mathrm{C}}_1{\mathrm{C}}_2{\mathrm{V}}_1^2-{\mathrm{C}}_2^2{\mathrm{V}}_2^2}{\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}\right] \\ =\frac{1}{2\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}\left[{\mathrm{C}}_1{\mathrm{C}}_2{\mathrm{V}}_1^2+{\mathrm{C}}_1{\mathrm{C}}_2{\mathrm{V}}_2^2+2{\mathrm{C}}_1{\mathrm{C}}_2{\mathrm{V}}_1{\mathrm{V}}_2\right] \\ =\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{2\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}{\left({\mathrm{V}}_1+{\mathrm{V}}_2\right)}^2 \end{gathered}$$

For a solid charged non-conducting sphere, for an inside point at distance r from center

$$\begin{gathered} V=\frac{kQ}{2R^3}(3R^2-r^2) \\ At center, r=0 \\ {V}_{center}=\frac{3kQ}{2R} \end{gathered}$$

$$\begin{gathered} At surface r=R, V_s=\frac{kQ}{R} \\ \therefore V_{center}=\frac{3V_s}{2} \\ =\frac{3}{2}\times 20 = 30V \end{gathered}$$

$$\begin{gathered} C_1=\frac{k{\epsilon }_{0}A}{d}\times 4 \\ {C}_2=\frac{k{\epsilon }_{0}A}{3d}\times 4 \\ {C}_{eq}=\frac{C_1{C}_2}{C_1+C_2} \\ {C}_{eq}=\frac{4KC}{3K+1} \end{gathered}$$