2 Marks Questions

Question 12 Marks

A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10cm and of radii 2mm and 4mm:

Calculate the capacitance.
Another capacitor of the same length is constructed with cylinders of radii 4mm and 8mm. Calculate the capacitance.

Answer

$\text{C}=\frac{2\in_0\text{L}}{\text{ln}\big(\frac{\text{R}_2}{\text{R}_1}\big)}=\frac{\text{e}\times3.14\times8.85\times10^{-2}\times10^{-1}}{\text{ln}2}$ $[\text{ln 2}=0.6932]$

$=80.17\times10^{-13}$

$\Rightarrow8\text{PF}$

Same as $\frac{\text{R}_2}{\text{R}_1}$ will be same.

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Question 22 Marks

A point charge Q is placed at the origin. Find the electrostatic energy stored outside the sphere of radius R centred at the origin.

Answer

Charge = Q

Radius of sphere = R

$\therefore$ Capacitance of the sphere $=\text{C}=4\pi\in_0\text{R}$

Energy $=\frac{1}{2}\frac{\text{Q}^2}{\text{C}}=\frac{1}{2}\frac{\text{Q}^2}{4\pi\in_0\text{R}}=\frac{\text{Q}^2}{8\pi\in_0\text{R}}$

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Question 32 Marks

The plates of a parallel-plate capacitor are made of circular discs of radii 5.0cm each. If the separation between the plates is 1.0mm, what is the capacitance?

Answer

$\text{A}=\pi\text{r}^2=25\pi\text{m}^2$

$\text{d}=0.1\text{cm}$

$\text{c}=\frac{\in_0\text{A}}{\text{d}}$

$=\frac{8.854\times10^{-12}\times25\times3.14}{0.1}$

$=6.95\times10^{-5}\mu\text{F}$

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Question 42 Marks

Two conducting spheres of radii R₁ and R₂ are kept widely separated from each other. What are their individual capacitances? If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series-parallel connections.

Answer

$\text{C}=\frac{\text{q}}{\text{V}},$ Now $\text{V}=\frac{\text{kq}}{\text{R}}$

So, $\text{C}_1=\frac{\text{q}}{\Big(\frac{\text{Kq}}{\text{R}_1}\Big)}=\frac{\text{R}_1}{\text{K}}=4\pi\in_0\text{R}_1$

Similarly, $\text{c}_2=4\pi\in_0\text{R}_2$

The combination is necessarily parallel.

Hence $\text{C}_{\text{eq}}=4\pi\in_0\text{R}_1+4\pi\in_0\text{R}_2=4\pi\in_0(\text{R}_1+\text{R}_2)$

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Question 52 Marks

Find the equivalent capacitance of the syatem shown in figure between the points a and b.

Answer

C_eq between a & b

$=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}+\text{C}_3+\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}$

$=\text{C}_3+\frac{2\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}$ $(\therefore$ The three are parallel$)$

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Question 62 Marks

A 100pF capacitor is charged to a potential difference of 24V. It is connected to an uncharged capacitor of capacitance 20pF. What will be the new potential difference across the 100pF capacitor?

Answer

Given that

C = 100PF = 100 × 10^-12F

C_eq = 20PF = 20 × 10^-12F

V = 24v

q = 24 × 100 × 10^-12 = 24 × 10^-10

q₂ = ?

Let q₁ = The new charge 100PF

V₁ = The Voltage.

Let the new potential is V₁

After the flow of charge, potential is same in the two capacitor

$\text{V}_1=\frac{\text{q}_2}{\text{C}_2}=\frac{\text{q}_1}{\text{C}_1}$

$=\frac{\text{q}-\text{q}_1}{\text{C}_2}=\frac{\text{q}_1}{\text{C}_1}$

$=\frac{24\times10^{-10}-\text{q}_1}{24\times10^{-12}}=\frac{\text{q}_1}{100\times10^{-12}}$

$=24\times10^{-10}-\text{q}_1=\frac{\text{q}_1}{5}$

$=6\text{q}_1=120\times10^{-10}$

$=\text{q}_1=\frac{120}{6}\times10^{-10}=20\times10^{-10}$

$\therefore\text{V}_1=\frac{\text{q}_1}{\text{C}_1}=\frac{20\times10^{-10}}{100\times10^{-12}}=20\text{v}$

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