Physics 5 Marks Questions

Plate area A = 25cm² = 2.5 × 10^-3m

Separation d = 2mm = 2 × 10^-3m

Potential v = 12v

1. We know

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

$=\frac{8.85\times10^{-12}\times2.5\times10^{-3}}{2\times10^{-3}}$

$=11.06\times10^{-12}\text{F}$

$\text{C}=\frac{\text{q}}{\text{v}}$

$\Rightarrow11.06\times10^{-12}=\frac{\text{q}}{12}$

$\Rightarrow\text{q}_1=1.32\times10^{-10}\text{C}.$

2. Then d = decreased to 1mm

$\therefore\ \text{d}=1\text{mm}=1\times10^{-3}\text{m}$

$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{\text{q}}{\text{v}}$

$=\frac{8.85\times10^{-12}\times2.5\times10^{-3}}{1\times10^{-3}}=\frac{2}{12}$

$\Rightarrow\text{q}_2=8.85\times2.5\times12\times10^{-12}$

$\Rightarrow\text{q}_2=2.65\times10^{-10}\text{C}.$

$\therefore$ The extra charge given to plate = (2.65 - 1.32) × 10^-10 = 1.33 × 10^-10C.

The capacitance of the outer sphere $=2.2\mu\text{F}$

$\text{C}=2.2\mu\text{F}$

Potential, V = 10v

Let the charge given to individual cylinder = q.

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

$\Rightarrow\text{q}=\text{CV}=2.2\times10=22\mu\text{F}$

$\therefore$ The total charge given to the inner cylinder $=22+22=44\mu\text{F}$

1. $\text{C}_1=5\mu\text{f}$

$\text{V}_1=24\text{V}$

$\text{q}_1=\text{C}_1\text{V}_1=5\times24=120\mu\text{c}$

and $\text{C}_2=6\mu\text{f}$

$\text{V}_2=\text{R}$

$\text{q}_2=\text{C}_2\text{V}_2=6\times12=72$

$\therefore$ Energy stored on first capacitor

$\text{E}_1=\frac{1}2{}\frac{\text{q}_1^2}{\text{C}_1}=\frac{1}{2}\times\frac{(120)^2}{2}=1440\text{J}=1.44\text{mJ}$

Energy stored on 2^nd capacitor

$\text{E}_2=\frac{1}2{}\frac{\text{q}_2^2}{\text{C}_1}=\frac{1}{2}\times\frac{(72)^2}{6}=432\text{J}=4.32\text{mJ}$

2. C₁V₁

C₂V₂

Let the effective potential = V

$\text{V}=\frac{\text{C}_1\text{V}_1-\text{C}_2\text{V}_2}{\text{C}_1+\text{C}_2}=\frac{120-72}{5+6}=4.36$

The new charge $\text{C}_1\text{V}=5\times4.36=21.8\mu\text{c}$

and $\text{C}_2\text{V}=6\times4.36=26.2\mu\text{c}$

3. $\text{U}_1=\Big(\frac{1}{2}\Big)\text{C}_1\text{V}^2$

$\text{U}_2=\Big(\frac{1}{2}\Big)\text{C}_2\text{V}^2$

$\text{U}_{\text{f}}=\Big(\frac{1}{2}\Big)\text{V}^2(\text{C}_1+\text{C}_2)$

$=\Big(\frac{1}{2}\Big)(4.36)^2(5+6)$

$=104.5\times10^{-6}\text{J}=0.1045\text{mJ}$

But $\text{U}_{\text{i}}=1.44+0.433=1.873$

$\therefore$ The loss in KE = 1.873 - 0.1045 = 1.7687 = 1.77mJ

1. Capacitor $=\frac{4\times8}{4+8}=\frac{8}{3}\mu\text{F}$

and $=\frac{6\times3}{6+3}=2\mu\text{F}$

1. The charge on the capacitance $\frac{8}{3}\mu\text{F}$

$\therefore\text{Q}=\frac{8}{3}\times50=\frac{400}{3}$

$\therefore$ The potential at $4\mu\text{F}=\frac{400}{3\times4}=\frac{100}{3}$

at $8\mu\text{F}=\frac{400}{3\times8}=\frac{100}{6}$

The Potential difference $=\frac{100}{3}-\frac{100}{6}=\frac{50}{3}\mu\text{V}$

2. Hence the effective charge at $2\mu\text{F}=50\times2=100\mu\text{F}$

$\therefore$ Potential at $3\mu\text{F}=\frac{100}{3};$ Potential at $6\mu\text{F}=\frac{100}{6}$

$\therefore$ Difference $=\frac{100}{3}-\frac{100}{6}=\frac{50}{3}\mu\text{V}$

$\therefore$ The potential at C & D is $\frac{50}{3}\mu\text{V}$

2. $\therefore\frac{\text{P}}{\text{Q}}=\frac{\text{R}}{\text{S}}=\frac{1}2{}=\frac{1}{2}=$ It is balanced. So, from it is cleared that the wheat star bridge balanced.

So, the potential at the point C & D are same. So, no current flow through the point C & D.

So, if we connect another capacitor at the point C & D the charge on the capacitor is zero.

1. Area = A

Separation = d

$\text{C}_1=\frac{\in_0\text{Ak}_1}{\frac{\text{d}}{2}}$

$\text{C}_2=\frac{\in_0\text{Ak}_2}{\frac{\text{d}}{2}}$

$\text{C}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\frac{\frac{2\in_0\text{Ak}_1}{\text{d}}\times\frac{2\in_0\text{Ak}_2}{\text{d}}}{\frac{2\in_0\text{Ak}_1}{\text{d}}+\frac{2\in_0\text{Ak}_2}{\text{d}}}$

$=\frac{\frac{(2\in_0\text{A})^2\text{k}_1\text{k}_2}{\text{d}^2}}{(2\in_0\text{A})\frac{\text{k}_1\text{d+}\text{k}_2\text{d}}{\text{d}^2}}=\frac{2\text{k}_1\text{k}_2\in_0\text{A}}{\text{d}(\text{k}_1+\text{k}_2)}$

2. $\frac{1}{\text{C}}=\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}$

$=\frac{1}{\frac{3\in_0\text{Ak}_1}{\text{d}}}+\frac{1}{\frac{3\in_0\text{Ak}_2}{\text{d}}}+\frac{1}{\frac{3\in_0\text{Ak}_3}{\text{d}}}$

$=\frac{\text{d}}{3\in_0\text{A}}\Big[\frac{1}{\text{k}_1}+\frac{1}{\text{k}_2}+\frac{1}{\text{k}_3}\Big]$

$=\frac{\text{d}}{3\in_0\text{A}}\Big[\frac{\text{k}_2\text{k}_3+\text{k}_1\text{k}_3+\text{k}_1\text{k}_2}{\text{k}_1\text{k}_2\text{k}_3}\Big]$

$\therefore\text{C}=\frac{3\in_0\text{A}\text{k}_1\text{k}_2\text{k}_3}{\text{d}(\text{k}_1\text{k}_2+\text{k}_2\text{k}_3+\text{k}_1\text{k}_3)}$

3. $\text{C}=\text{C}_1+\text{C}_2$

$=\frac{\in_0\frac{\text{A}}{2}\text{k}_1}{\text{d}}+\frac{\in_0\frac{\text{A}}{2}\text{k}_2}{\text{d}}=\frac{\in_0\text{A}}{2\text{d}}(\text{k}_1+\text{k}_2)$

In the figure the three capacitors are arranged in parallel.

All have same surface area $=\text{a}=\frac{\text{A}}{3}$

First capacitance $\text{C}_1=\frac{\in_0\text{A}}{3\text{d}}$

Second capacitance $\text{C}_2=\frac{\in_0\text{A}}{3(\text{b}+\text{d})}$

Third capacitance $\text{C}_3=\frac{\in_0\text{A}}{3(2\text{b}+\text{d})}$

$\text{C}_{\text{eq}}=\text{C}_1+\text{C}_2+\text{C}_3$

$=\frac{\in_0\text{A}}{3\text{d}}+\frac{\in_0\text{A}}{3(\text{b}+\text{d})}+\frac{\in_0\text{A}}{3(2\text{b}+\text{d})}$

$=\frac{\in_0\text{A}}{3}\Big(\frac{1}{\text{d}}+\frac{1}{\text{b}+\text{d}}+\frac{2}{\text{2b}+\text{d}}\Big)$

$=\frac{\in_0\text{A}}{3}\Big(\frac{(\text{b}+\text{d})(2\text{b}+\text{d})+(2\text{b}+\text{d})\text{d}+(\text{b}+\text{d})\text{d}}{\text{d}(\text{b}+\text{d})(2\text{b}+\text{d})}\Big)$

$=\frac{\in_0\text{A}\big(3\text{d}^2+6\text{bd}+2\text{b}^2\big)}{3\text{d}(\text{b}+\text{d})(2\text{b}+\text{d})}$

1. When charge of $1\mu\text{C}$ is introduced to the B plate, we also get $0.5\mu\text{C}$ charge on the upper surface of Plate ‘A’.
2. Given $\text{C}=50\mu\text{F}=50\times10^{-9}\text{F}=5\times10^{-8}\text{F}$

Now charge $=0.5\times10^{-6}\text{C}$

$\text{V}=\frac{\text{q}}{\text{C}}=\frac{5\times10^{-7}\text{C}}{5\times10^{-8}\text{F}}=10\text{V}$

1. Before reconnection

$\text{C}=100\mu\text{f}$

$\text{V}=24\text{V}$

$\text{q}=\text{CV}=2400\mu\text{C}$ (Before reconnection)

After connection

When $\text{C}=100\mu\text{f}$

$\text{V}=12\text{V}$

$\text{q}=\text{CV}=1200\mu\text{C}$ (After connection)

2. C = 100, V = 12V

$\therefore$ q = CV = 1200v

3. We know $\text{V}=\frac{\text{W}}{\text{q}}$

W = vq = 12 × 1200 = 14400 J = 14.4 mJ

The work done on the battery.

4. Initial electrostatic field energy $\text{U}_{\text{i}}=\Big(\frac{1}{2}\Big)\text{CV}_1^2$

Final Electrostatic field energy $\text{U}_{\text{f}}=\Big(\frac{1}{2}\Big)\text{CV}_2^2$

$\therefore$ Decrease in Electrostatic

Field energy $=\Big(\frac{1}2{}\Big)\text{CV}_1^2-\Big(\frac{1}2{}\Big)\text{CV}_2^2$

$=\Big(\frac{1}{2}\Big)\text{C}(\text{V}_1^2-\text{V}_2^2)$

$=\Big(\frac{1}{2}\Big)\times100(576-144)=21600\text{J}$

$\therefore$ Energy = 21600j = 21.6mJ

5. After reconnection

$\text{C}=100\mu\text{c},\ \text{V}=12\text{v}$

$\therefore$ The energy appeared $=\Big(\frac{1}{2}\Big)\times100\times144$

$=7200\text{J}=7.2\text{mJ}$ 

This amount of energy is developed as heat when the charge flow through the capacitor.

The acceleration of electron $\text{a}_{\text{e}}=\frac{\text{qeme}}{\text{Me}}$

The acceleration of proton $=\frac{\text{qpe}}{\text{Mp}}=\text{ap}$

The distance travelled by proton $\text{X}=\frac{1}{2}\text{apt}^2\ \dots(1)$

The distance travelled by electron ...(2)

From (1) and (2)

$\Rightarrow2-\text{X}=\frac{1}{2}\text{a}_{\text{c}}\text{t}^2$

$\text{x}=\frac{1}{2}\text{a}_{\text{c}}\text{t}^2$

$\Rightarrow\frac{\text{x}}{2-\text{x}}=\frac{\text{a}_{\text{p}}}{\text{a}_{\text{c}}}=\frac{\big(\frac{\text{q}_{\text{p}\text{E}}}{\text{M}_{\text{p}}}\big)}{\big(\frac{\text{q}_{\text{c}\text{F}}}{\text{M}_{\text{c}}}\big)}$

$=\frac{\text{x}}{2-\text{x}}=\frac{\text{M}_{\text{c}}}{\text{M}_{\text{p}}}=\frac{9.1\times10^{-31}}{1.67\times10^{-27}}$

$=\frac{9.1}{1.67}\times10^{-4}=5.449\times10^{-4}$

$\Rightarrow\text{x}=10.898\times10^{-4}-5.449\times10^{-4}\text{x}$

$\Rightarrow\text{x}=\frac{10.898\times10^{-4}}{1.0005449}=0.001089226$

Consider an elemental capacitor of with dx our at a distance ‘x’ from one end. It is constituted of two capacitor elements of dielectric constants k₁ and k₂ with plate separation $\text{x}\tan\phi$ and $\text{d}-\text{x}\tan\phi$ respectively in series.

$\frac{1}{\text{dcR}}=\frac{1}{\text{dc}_1}+\frac{1}{\text{dc}_2}=\frac{\text{x}\tan\phi}{\in_0\text{k}_2(\text{bdx})}+\frac{\text{d}-\text{x}\tan\phi}{\in_0\text{k}_1(\text{bdx})}$

$\text{dcR}=\frac{\in_0\text{bdx}}{\frac{\text{x}\tan\phi}{\text{k}_2}+\frac{(\text{d}-\text{x}\tan\phi)}{\text{k}_1}}$

$\text{C}_{\text{R}}=\in_0\text{bk}_1\text{k}_2\int\frac{\text{dx}}{\text{k}_2\text{d}+(\text{k}_1+\text{k}_2)\text{x}\tan\phi}$

$=\frac{\in_0\text{bk}_1\text{k}_2}{\tan\phi(\text{k}_1-\text{k}_2)}[\log_{\text{e}}\text{k}_2\text{d}+(\text{k}_1+\text{k}_2)\text{x}\tan\phi]\text{a}$

$=\frac{\in_0\text{bk}_1\text{k}_2}{\tan\phi(\text{k}_1-\text{k}_2)}[\log_{\text{e}}\text{k}_2\text{d}+(\text{k}_1+\text{k}_2)\text{a}\tan\phi-\log_{\text{e}}\text{k}_2\text{d}]$

$\therefore\tan\phi=\frac{\text{c}}{\text{a}}$ and $\text{A}=\text{a}\times\text{a}$

$\text{C}_{\text{R}}=\frac{\in_0\text{ak}_1\text{k}_2}{\frac{\text{d}}{\text{a}}(\text{k}_1-\text{k}_2)}$ $\Big[\log_{\text{e}}\Big(\frac{\text{k}_1}{\text{k}_2}\Big)\Big]$

$\text{C}_{\text{R}}=\frac{\in_0\text{a}^2\text{k}_1\text{k}_2}{\text{d}(\text{k}_1-\text{k}_2)}$ $\Big[\log_{\text{e}}\Big(\frac{\text{k}_1}{\text{k}_2}\Big)\Big]$

$\text{C}_{\text{R}}=\frac{\in_0\text{a}^2\text{k}_1\text{k}_2}{\text{d}(\text{k}_1-\text{k}_2)}$ $\text{ln}\frac{\text{k}_1}{\text{k}_2}$

1. The Electric field is along x-direction

Thus potential difference between,

$\text{A}=(0, 0);\ \text{B}=(4, 2)$

$\Rightarrow\delta\text{V}=-\text{E}\times\delta\text{x}$

$=-20\times(40)$

$=-80\text{V}$

Potential energy (U_B - U_A) between the points $=\delta\text{V}\times\text{q}$

$=-80\times(-2)\times10^{-4}$

$=160\times10^{-4}$

$=0.016\text{J}$

2. $\text{A}=(4\text{m},2\text{m});\ \text{B}=(6\text{m},5\text{m})$

$\Rightarrow\delta\text{V}=-\text{E}\times\delta\text{x}$

$=-20\times2$

$=-40\text{V}$

Potential energy (U_B - U_A) between the points $=\delta\text{V}\times\text{q}$

$=-40\times(-2)\times10^{-4})$

$=80\times10^{-4}$

$=0.008\text{J}$

3. $\text{A}=(0,0);\ \text{B}=(6\text{m}, 5\text{m})$

$\Rightarrow\delta\text{V}=-\text{E}\times\delta\text{x}$

$=-20\times6$

$=-120\text{V}$

Potential energy (U_B - U_A) between the points A and B

$\Rightarrow\delta\text{V}\times\text{q}$

$=-120\times(-2\times10^{-4})$

$=240\times10^{-4}$

$=0.024\text{J}$