Numerical Problems-2 - Chapter 1 Electrostatics 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Numerical Problems
Question 1.
Electrons are caused to fall through a potential difference of 1500 volts. If they were initially at rest. Then calculate their final speed.
Solution:
The electrical potential energy is converted into kinetic energy. If $\mathrm{v}$ is the final speed then
$
\begin{aligned}
\frac{1}{2} m v^2 & =e \mathrm{~V} \\
v & =\sqrt{\frac{2 e \mathrm{~V}}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1500}{9.1 \times 10^{-31}}}=2.3 \times 10^7 \mathrm{~ms}^{-1}
\end{aligned}
$
Question 2.
Small mercury drops of the same size are charged to the same potential $\mathrm{V}$. If $\mathrm{n}$ such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let $\mathrm{r}$ be the radius of a small drop and $\mathrm{R}$ that of the large drop. Then, since the volume remains conserved,
$
\begin{aligned}
& \frac{1}{2} \pi R^2=\frac{4}{3} \pi R^3 n \\
& \Rightarrow R^3=r^3 n \\
& R=r^3(n)^{1 / 3}
\end{aligned}
$
Further, since the total charge remains conserved, we have, using $Q=C V$
$\mathrm{C}_{\text {large }} \mathrm{V}=\mathrm{n} \mathrm{C}_{\text {small }} \mathrm{v}$
Where $V$ is the potential of the large drop.
$4 \pi \varepsilon_0 \mathrm{RV}=\mathrm{n}\left(4 \pi \varepsilon_0 \mathrm{r}\right) \mathrm{v}$
$
\begin{aligned}
& \mathrm{V}=\frac{n r v}{R}=\frac{n r v}{r(n)^{1 / 3}} \\
& \mathrm{~V}=\mathrm{vn}^{2 / 3}
\end{aligned}
$

Question 3.
Two particles having charges $Q_1$ and $Q_2$ when kept at a certain distance, exert a force $F$ on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles.
Solution:
$
\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}_1 \mathrm{Q}_2}{r^2}
$
If the distance is educed by half and two particles of charges are doubled.
$
\begin{aligned}
\mathrm{F}^{\prime} & =\frac{1}{4 \pi \varepsilon_0} \frac{\left(2 \mathrm{Q}_1\right)\left(2 \mathrm{Q}_2\right)}{(r / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{4 \mathrm{Q}_1 \mathrm{Q}_2}{\left(r^2 / 4\right)^2} \\
\mathrm{~F}^{\prime} & =\frac{1}{4 \pi \varepsilon_0} \frac{16\left(\mathrm{Q}_1 \mathrm{Q}_2\right)}{r^2}=16\left[\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}_1 \mathrm{Q}_2}{r^2}\right] \\
\mathrm{F}^{\prime} & =16 \mathrm{~F}
\end{aligned}
$
Question 4.
Two charged spheres, separated by a distance $d$, exert a force $F$ on each other. If they are immersed in a liquid of dielectric constant 2 , then what is the force.
Solution:
Force between the charges (vacuum)
$
\mathrm{F}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2}
$

Force between the charges (medium)
$
\begin{aligned}
& \mathrm{F}^{\prime}=\frac{1}{4 \pi \varepsilon} \frac{q_1 q_2}{d^2} \quad\left[\varepsilon=\varepsilon_0 \varepsilon_r\right] \\
& \mathrm{F}^{\prime}=\frac{1}{4 \pi \varepsilon_0 \varepsilon_r} \frac{q_1 q_2}{d^2}=\frac{1}{4 \pi \varepsilon_0(2)} \frac{q_1 q_2}{d^2}=\frac{1}{2}\left[\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2}\right] \\
& \mathrm{F}^{\prime}=\frac{\mathrm{F}}{2}
\end{aligned}
$
Question 5.
Find the force of attraction between the plates of a parallel plate capacitor.
Solution:
Let $\mathrm{d}$ be the distance between the plates. Then the capacitor is
$
\mathrm{C}=\frac{\varepsilon_0 A}{d}
$
Energy stored in a capacitor,
$
\mathrm{U}=\frac{q^2}{2 \mathrm{C}}=\frac{q^2 \cdot d}{2 \varepsilon_0 \mathrm{~A}}
$
Energy magnitude of the force is,
$
\begin{aligned}
|\mathrm{F}| & =\frac{d \mathrm{U}}{d x}=\frac{d}{d x}\left[\frac{q^2 x}{2 \varepsilon_0 \mathrm{~A}}\right] \quad[x=d] \\
& =\frac{q^2}{2 \varepsilon_0 \mathrm{~A}}\left[\frac{d}{d x}(x)\right] \\
\mathrm{F} & =\frac{q^2}{2 \varepsilon_0 \mathrm{~A}}
\end{aligned}
$