Numerical Problems

Question 15 Marks

Find the force of attraction between the plates of a parallel plate capacitor.

Answer

Let $d$ be the distance between the plates. Then the capacitor is
$C =\frac{\varepsilon_0 A}{d}$
Energy stored in a capacitor,
$U =\frac{q^2}{2 C }=\frac{q^2 \cdot d}{2 \varepsilon_0 A }$
Energy magnitude of the force is,
$| F |=\frac{d U }{d x}=\frac{d}{d x}\left[\frac{q^2 x}{2 \varepsilon_0 A }\right] \quad[x=d]$
$=\frac{q^2}{2 \varepsilon_0 A }\left[\frac{d}{d x}(x)\right]$
$F =\frac{q^2}{2 \varepsilon_0 A }$

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Question 25 Marks

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.

Answer

Force between the charges (vacuum)
$F =\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{d^2}$
Force between the charges (medium)
$F ^{\prime}=\frac{1}{4 \pi \varepsilon} \frac{q_1 q_2}{d^2} \quad\left[\varepsilon=\varepsilon_0 \varepsilon_r\right] $
$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] $
$F ^{\prime}=\frac{ F }{2}$

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Question 35 Marks

Two particles having charges Q1 and Q2 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.

Answer

$
F =\frac{1}{4 \pi \varepsilon_0} \frac{ Q _1 Q _2}{r^2}
$
If the distance is educed by half and two particles of charges are doubled.
$\begin{aligned} & F ^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{\left(2 Q _1\right)\left(2 Q _2\right)}{(r / 2)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{4 Q _1 Q _2}{\left(r^2 / 4\right)^2} \\ & F ^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{16\left( Q _1 Q _2\right)}{r^2}=16\left[\frac{1}{4 \pi \varepsilon_0} \frac{ Q _1 Q _2}{r^2}\right] \\ & F ^{\prime}=16 F \end{aligned}$

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Question 45 Marks

Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.

Answer

Let $r$ be the radius of a small drop and $R$ that of the large drop. Then, since the volume remains conserved,
$\frac{1}{2} \pi R^2=\frac{4}{3} \pi R^3 n$
$\Rightarrow R^3=r^3 n$
$R=r^3(n)^{1 / 3}$
Further, since the total charge remains conserved, we have, using $Q=$ CV
$C _{\text {large }} V = n C _{\text {small }} V$
Where $V$ is the potential of the large drop.
$4 \pi \varepsilon_0 RV = n \left(4 \pi \varepsilon_0 r \right) v$
$V =\frac{n r v}{R}=\frac{n r v}{r(n)^{1 / 3}}$
$V = vn ^{2 / 3}$

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Question 55 Marks

Electrons are caused to fall through a potential difference of 1500 volts. If they were initially at rest. Then calculate their final speed.

Answer

The electrical potential energy is converted into kinetic energy. If v is the final speed then
$
\begin{aligned}
\frac{1}{2} m v^2 & =e V _
v & =\sqrt{\frac{2 e V }{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 1500}{9.1 \times 10^{-31}}}=2.3 \times 10^7 ms ^{-1}
\end{aligned}
$

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Physics Numerical Problems

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