Physics 3 Marks Question

$\frac{1}{4\pi\in_0}\frac{\text{q}_1}{\text{R}_1}=\frac{1}{4\pi\in_0}\frac{\text{q}_2}{\text{R}_2}$

$\Rightarrow\ \frac{\text{R}_1}{\in_0}\frac{\text{q}}{4\pi\text{R}_1^2}=\frac{\text{R}_2}{\in_0}\frac{\text{q}_2}{4\pi\text{R}_2^2}$

$\text{U}_{\text{initial}}=\frac{1}{4\pi\varepsilon_{0}}\Big[\frac{\text{Q}(-2\text{Q})}{\text{l}}+\frac{\text{Q}(-3\text{Q})}{\text{l}}+\frac{2\text{Q}(3\text{Q})}{\text{l}}\Big]$

$=\frac{1}{4\pi\varepsilon_0}\frac{1\text{Q}^2}{\text{l}}.$

$\text{U}_{\text{initial}}=\frac{1}{4\pi\varepsilon_0}\bigg[\frac{\text{Q}(-2\text{Q})}{\frac{\text{l}}{2}}+\frac{\text{Q}(-3\text{A})}{\frac{\text{l}}{2}}+\frac{2\text{Q}(3\text{Q})}{\frac{\text{l}}{2}}\bigg]$

$=\frac{1}{4\pi\varepsilon_0}\frac{2\text{Q}^2}{\text{l}}.$

$\text{W}=\text{U}_{\text{final}}-\text{U}_{\text{initial}}=-\frac{1}{4\pi\varepsilon_0}\frac{2\text{Q}^2}{\text{l}}$

$\text{R}_\text{eff}=10+20=30\Omega$

$\text{i}=\frac{2}{30}=\frac{1}{15}\text{A}$

$\therefore$ Electrostatic potential energy of charges,

$\text{U}=\frac{1}{4\pi\varepsilon_0}\frac{\text{q}_1\text{q}_2}{\text{x}}$

$\frac{9\times10^9\big(5\times10^{-9}\big)\big(-2\times10^{-9}\big)}{0.18}=5\times10-7\text{J}$

1. Electric field is radial,
   
2. The electric field $\text{E}\propto\frac{1}{\text{R}^2}$
   

$\text{V}=\text{10v}$

$\text{C}_{\text{eq}}=\text{C}_1+\text{C}_2$ $[\therefore$ They are parallel$]$

$=5+6=11\mu\text{F}$

$\text{q}=\text{CV}=11\times10=110\mu\text{C}$

For a single charge the potential is given by $\text{V}=\frac{1}{4\pi\varepsilon_0}\frac{\text{q}}{\text{r}}$

This shows that V is constant if r is constant. Greater the radius smaller will be the potential. In the given figure, potential is increasing. This shows that the polarity of charge is negative (-q). The direction of electric field will be radially inward. The field lines are directed from higher to lower potential.

$\therefore$ Given that

Capacitance $=10\mu\text{F}$

Charge $=20\mu\text{C}$

$\therefore$ The effective charge $=\frac{20-0}{2}=10\mu\text{F}$

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

$\Rightarrow\text{V}=\frac{\text{q}}{\text{C}}=\frac{10}{10}=1\text{V}$

$\text{C}_1=8\mu\text{F},\ \text{C}_2=4\mu\text{F},\ \text{C}_3=4\mu\text{F}$

$\text{C}_{\text{eq}}=\frac{(\text{C}_2+\text{C}_3)\times\text{C}_1}{\text{C}_1+\text{C}_2+\text{C}_3}$

$=\frac{8\times8}{16}=4\mu\text{F}$

Since B & C are parallel & are in series with A.

So, $\text{q}_1=8\times6=48\mu\text{C}$

$\text{q}_2=4\times6=24\mu\text{C}$

$\text{q}_3=4\times6=24\text{C}\mu$

1. The potential due to positive charge is positive and due to negative charge, it is negative, so, is q₁ positive and q₂ is negative.

$\text{V}=\frac{1}{4\pi\varepsilon_0}\frac{\text{q}}{\text{r}}$

2. The graph between V and $\frac{1}{\text{r}}$ is a straight line passing through the origin with slope $\frac{\text{q}}{4\pi\varepsilon_0},$

As the magnitude of slope of the line due to charge q₂ is greater than that due to q₁, q₂ has larger magnitude.

Here three capacitors are formed

And each of $\text{A}=\frac{96}{\in_0}\times10^{-12}\text{f.m.}$

$\text{d}=4\text{mm}=4\times10^{-3}\text{m}$

$\therefore$ Capacitance of a capacitor

$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{\in_0\frac{96\times10^{-12}}{\in_0}}{4\times10^{-3}}=24\times10^{-9}\text{F}$

$\therefore$ As three capacitor are arranged is series

So, $\text{C}_\text{eq}=\frac{\text{C}}{\text{q}}=\frac{24\times10^{-9}}{3}=8\times10^{-9}$

$\therefore$ The total charge to a capacitor = 8 × 10^-9 × 10 = 8 × 10^-8c

$\therefore$ The charge of a single Plate = 2 × 8 × 10^-8 = 16 × 10^-8 = 0.16 × 10^-6 = 0.16μc.

$\therefore$ The net potential difference = 10L.

$\text{V}=\frac{\text{q}_1\text{q}_2}{4\pi\in_0\text{d}_1}+\frac{\text{q}_2\text{q}_3}{4\pi\in_0\text{d}_1}$

$\therefore\text{C}_1=4\mu\text{F},\ \text{C}_2=6\mu\text{F},\ \text{V}=20\text{V}$

Eq. capacitor $\text{C}_{\text{eq}}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\frac{4\times6}{4+6}=2.4$

$\therefore$ The Eq Capacitance $\text{C}_{\text{eq}}=2.5\mu\text{F}$

$\therefore$ The energy supplied by the battery to each plate

$\text{E}=\Big(\frac{1}{2}\Big)\text{CV}^2$

$=\Big(\frac{1}{2}\Big)\times2.4\times20^2=480\mu\text{J}$

$\therefore$ The energy supplies by the battery to capacitor $=2\times480=960\mu\text{J}$

$\text{C}_1=2\mu\text{F},\ \text{C}_2=4\mu\text{F},\ \text{C}_3=6\mu\text{F}$

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

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

$=2+4+6=12\mu\text{F}$

$=12\times10^{-6}\text{F}$

$\text{q}_1=12\times2=24\mu\text{C}$

$\text{q}_2=12\times4=48\mu\text{C}$

$\text{q}_3=12\times6=72\mu\text{C}$