Capacitor

Using $C_{e q}=\frac{\varepsilon_0 A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}}$

here $C_0=\frac{\varepsilon_0 A}{d}, t_1=\frac{3 d}{8}, t_2=\frac{d}{2}, t_3=\frac{d}{8}$

$K_1=K_1, \quad K_2=\frac{K_1}{1.25} \text { and } K_3=1$

Given $C_{\text {eq }}=2 C_0$

$\begin{aligned} & \Rightarrow 2 C_0=\frac{\varepsilon_0 A}{\frac{3 d}{8 k_1}+\frac{d \times 1.25}{2 k_1}+\frac{d}{8}} \\ & \Rightarrow \frac{2 \varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{\frac{3 d}{8 k_1}+\frac{d}{2 k_1} \times \frac{5}{4}+\frac{d}{8}} \\ & \Rightarrow 2=\frac{1}{\frac{3}{8 k_1}+\frac{5}{8 k_1}+\frac{1}{8}} \Rightarrow k_1=\frac{8}{3}=2.66 \end{aligned}$

The electrostatic energy stored in a capacitor can be calculated using the formula:

$ U = \frac{1}{2} C V^2 $

Where:

• $ U $ is the stored energy in joules (J)
• $ C $ is the capacitance in farads (F)
• $ V $ is the voltage in volts (V)

Given:

• $ C = 12 \mathrm{~pF} $ (which is $ 12 \times 10^{-12} $ F)
• $ V = 50 \mathrm{~V} $

Substitute the values into the formula:

$ U = \frac{1}{2} \times 12 \times 10^{-12} \times (50)^2 $

Calculate the values inside the parentheses first:

$ 50^2 = 2500 $

Now substitute this back into the equation:

$ U = \frac{1}{2} \times 12 \times 10^{-12} \times 2500 $

Perform the multiplication:

$ U = \frac{1}{2} \times 30000 \times 10^{-12} $

Which simplifies to:

$ U = 15000 \times 10^{-12} $

Convert this to nanojoules (nJ) by recognizing that $ 1 \mathrm{~nJ} = 10^{-9} \mathrm{~J} $:

$ U = 15 \mathrm{~nJ} $

Thus, the electrostatic energy stored in the capacitor is:

Option A: 15 nJ

The correct answer is Option B: the geometry of the capacitor. Here's why:

Capacitance (denoted by *C*) is a fundamental property of a capacitor that describes its ability to store electrical energy. It's defined by the ratio of the charge stored on the capacitor (q) to the potential difference across its plates (V):

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

While the formula seems to suggest that capacitance depends on both charge and voltage, the reality is that the geometry of the capacitor determines its capacitance. Here's a breakdown:

• Charge (q): The charge stored on a capacitor is directly proportional to the applied voltage. If you increase the voltage, you increase the charge stored. However, the capacitance itself remains constant for a given capacitor.


• Voltage (V): Similarly, the voltage across a capacitor is directly proportional to the charge stored. Increasing the charge increases the voltage, but again, the capacitance remains unchanged.


• Geometry: The geometry of a capacitor dictates how much electric field is created between its plates for a given charge. This electric field determines the potential difference between the plates. Here are some key factors:
  
  
  • Area of plates (A): Larger plates can hold more charge for a given voltage, resulting in higher capacitance.
  
  
  • Distance between plates (d): Smaller distances between plates create a stronger electric field, which leads to higher capacitance.
  
  
  • Dielectric material: The material between the plates (dielectric) affects the strength of the electric field and therefore the capacitance. A material with a higher dielectric constant increases the capacitance.

In summary, while charge and voltage are related to capacitance through the formula, they are not the determining factors. It's the capacitor's physical characteristics – its geometry – that ultimately determine its capacitance.

At steady state, capacitor will be completely charged and will not allow current to pass through it. The simplified circuit will be :

$i=\frac{V}{R} \Rightarrow i=\frac{10}{2+3}=2 \mathrm{~A}$

Given circuit is balanced Wheatstone bridge

$ \begin{aligned} C_{A B} & =1+1 \\ & =2 \mu F \end{aligned} $

Given $V^{\prime}=V=$ Constant

$\begin{aligned} \text{(i)}\quad & C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d} \\ & d^{\prime}< d \\ & C^{\prime}> C \end{aligned}$

Hence, final capacitance greater than initial capacitance,

$ \begin{aligned} \text { (ii) } \quad U^{\prime} & =\frac{1}{2} C^{\prime} V^2 \\ U & =\frac{1}{2} C V^2 \\ U^{\prime} & >U \end{aligned} $

Hence final energy is greater than initial energy

$ \begin{aligned} \text { (iii) }\quad \frac{Q^{\prime}}{V^{\prime}} & =C^{\prime} \text { and } \frac{Q}{V}=C \\ \frac{Q^{\prime}}{V^{\prime}} & \neq \frac{Q}{V} \end{aligned}$

(iv) Product of charge and voltage

$\begin{aligned} & X^{\prime}=Q^{\prime} V=C^{\prime} V^2 \\ & X=Q V=C V^2 \\ & X^{\prime}>X \end{aligned}$

$C_{eq}=5+15=20\mu F$

$ \begin{aligned} & \mathrm{Q}=\mathrm{CV} \\\\ & \begin{aligned} \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{C} \cdot \frac{\mathrm{dV}}{\mathrm{dt}} \Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}} & =\frac{\mathrm{I}}{\mathrm{C}}=\frac{2 \times 10^{-3}}{4 \times 10^{-6}} \\ & =\frac{10^3}{2}=500 \frac{\mathrm{V}}{\mathrm{s}} \end{aligned} \end{aligned} $

$\mathrm{C_{AB}=\frac{3\times6}{3+6}=2\mu F}$

For a capacitor of area A and distance between the plates as d, the capacitance (C) is

$C = {{A{ \in _0}} \over d}$

On doubling the distance and reducing area to half

${C_1} = {{{A \over 2}{ \in _0}} \over {2d}}$

$ \Rightarrow {C_1} = {{A{ \in _0}} \over {4d}}$

$ \Rightarrow {C_1} = {C \over 4}$

Here, ${1 \over {{C_1}}} + {1 \over {{C_2}}} = {1 \over 3}$ (Series combination)

While, ${C_1} + {C_2} = 16$ (Parallel combination)

Solving the above equations we get

${C_1} = 12\,\mu F$ and ${C_2} = 4\,\mu F$

q₁ = CV

= 900 $\times$ 10^$-$12 $\times$ 100

= 9 $\times$ 10^$-$8 C

$V = {{{C_1}{V_1} + {C_2}{V_2}} \over {{C_1} + {C_2}}}$

$ = {{9 \times {{10}^{ - 8}} + 0} \over {1800 \times {{10}^{ - 12}}}} = {{100} \over 2} = 50$ V

$U = {1 \over 2}({C_1} + {C_2}){V^2}$

$ = {1 \over 2} \times 1800 \times {10^{ - 12}} \times 50 \times 50$

$ = 225 \times {10^{ - 8}}$

$U = 2.25 \times {10^{ - 6}}$ J

At $t \rightarrow \infty$, capacitor

Works as a open circuit as shown in the Figure.

Therefore, current flow in the circuit,

$\begin{aligned} I_1 & =\frac{9}{(12+15) 10^3}=\frac{1}{3} \times 10^{-3} \mathrm{~A} \\ \therefore \quad V_0 & =I_1 \times 15 \times 10^3 \\ & =\frac{1}{3} \times 10^{-3} \times 15 \times 10^3=5 \mathrm{~V} \end{aligned}$

$\therefore$ Charge on capacitor of $5 \mu \mathrm{F}$,

$\begin{aligned} q_0 & =C V_0=5 \times 10^{-6} \times 5 \\ & =25 \times 10^{-6} \mathrm{C}=25 \mu \mathrm{C} \end{aligned}$

When switch is open, then capacitor starts discharging across $(5+15=20 \mathrm{k} \Omega)$ and instantaneous charge on capacitor is given by

$\begin{aligned} q & =q_0 e^{-t / R C} \\ & =25 e^{-\frac{1}{20 \times 10^3 \times 5 \times 10^{-6}}} \quad[\because t=1 \mathrm{~s}] \\ & =25 e^{-10} \mu \mathrm{C} \end{aligned}$

Given, capacitance of a capacitor

$\begin{aligned} & C=15 \mathrm{~nF}=15 \times 10^{-9} \mathrm{~F} \\ & \varepsilon_r=2.5 \end{aligned}$

Electric field, $E=30 \times 10^6 \mathrm{~V} / \mathrm{m}$

Potential difference, $V=30 \mathrm{~V}$

Area of plate $=$ ?

If $d$ be the distance between the plates, then

$d=\frac{V}{E}=\frac{30}{30 \times 10^6}=10^{-6} \mathrm{~m}$

Capacitance of capacitor,

$\begin{aligned} C & =\frac{\varepsilon_0 \varepsilon_r A}{d} \\ 15 \times 10^{-9} & =\frac{8.85 \times 10^{-12} \times 2.5 \times A}{10^{-6}} \\ \Rightarrow \quad A & =6.7 \times 10^{-4} \mathrm{~m}^2 \end{aligned}$

This combination forms a GP,

$s=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots$

Sum of infinite GP,

$\begin{aligned} & s=\frac{a}{1-r} \\ & s=\frac{1}{1-\frac{1}{2}} \\ & s=\frac{1}{\frac{1}{2}} \\ \therefore \quad & s=2 \end{aligned}$

Hence, capacitance of the combination

$C_{\infty}=2 \times 1 \mu \mathrm{F}=2 \mu \mathrm{F}$

Charge of plates remains same as there is no transfer of charge.

The two capacitors are in parallel.

The equivalent capacitance $C^{\prime}=C+\frac{C}{2}=\frac{3 C}{2}$

Work done in charging the equivalent capacitor is stored in the form of potential energy.

$\begin{aligned} W & =U=\frac{1}{2} C^{\prime} V^2 \\ & =\frac{1}{2} \times \frac{3 C}{2} \times V^2 \\ & =\frac{3}{4} C V^2 \end{aligned}$

Net charge on the parallel plate capacitor is $1 \mu \mathrm{C}$

Electric field near one of the Plates due to the other plate is $E=\frac{10^5}{2} \mathrm{~Vm}^{-1}$

Now, force is given by

$\begin{aligned} \qquad F & =q E \\ \text { Substituting } q & =1 \mu \mathrm{C}=1 \times 10^{-6} \mathrm{C} \\ \text { We get } \quad F & =0 \cdot 05 \mathrm{~N} \end{aligned}$