The space between the plates of a parallel plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $K_1=2, K_2=3$ and $K_3=5$ (as shown in figure). If new capacitance is $\frac{n}{3} C$ then the value of $n$ is $\_\_\_\_$ .
Explanation:
The capacitance of a parallel plate capacitor without any dielectric is given by:
$ C=\frac{\epsilon_0 A}{d} $
where A is the area of the plates and d is the separation between them.
Capacitor $\mathrm{C}_1$
Dielectric constant $\mathrm{K}_1=2$.
Thickness $\mathrm{d}_1=\frac{\mathrm{d}}{2}$.
Area $\mathrm{A}_1=\mathrm{A}$
So, the capacitance is;
$ C_1=\frac{K_1 \epsilon_0 A_1}{d_1}=\frac{2 \epsilon_0 A}{d / 2}=4\left(\frac{\epsilon_0 A}{d}\right)=4 C $
Capacitor $\mathrm{C}_2$
Dielectric constant $\mathrm{K}_2=3$.
Thickness $\mathrm{d}_2=\frac{\mathrm{d}}{2}$.
Area $\mathrm{A}_1=\mathrm{A} / 2$
So, the capacitance is;
$\mathrm{C}_2=\frac{\mathrm{K}_2 \epsilon_0 \mathrm{~A}_2}{\mathrm{~d}_2}=\frac{3 \epsilon_0(\mathrm{~A} / 2)}{\mathrm{d} / 2}=3\left(\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)=3 \mathrm{C}$
Capacitor $\mathrm{C}_3$
Dielectric constant $\mathrm{K}_3=5$.
Thickness $\mathrm{d}_3=\frac{\mathrm{d}}{2}$.
Area $\mathrm{A}_2=\mathrm{A} / 2$
So, the capacitance is;
$ \mathrm{C}_3=\frac{\mathrm{K}_3 \epsilon_0 \mathrm{~A}_3}{\mathrm{~d}_3}=\frac{5 \epsilon_0(\mathrm{~A} / 2)}{\mathrm{d} / 2}=5\left(\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}\right)=5 \mathrm{C} $
Capacitors $C_2$ and $C_3$ are in parallel.
$ C_{23}=C_2+C_3 $
$\Rightarrow $ $ C_{23}=3 C+5 C=8 C $
This combined section $\left(\mathrm{C}_{23}\right)$ is in series with the first capacitor $\left(\mathrm{C}_1\right)$.
$\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_{23}}$
$\Rightarrow $ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{4 \mathrm{C}}+\frac{1}{8 \mathrm{C}}$
$\Rightarrow $ $\frac{1}{C_{\mathrm{eq}}}=\frac{2+1}{8 C}=\frac{3}{8 C}$
$\Rightarrow $ $ C_{\mathrm{eq}}=\frac{8 C}{3}=\frac{n}{3} C \Rightarrow n=8 $
Therefore, the value of n is 8 . Hence, the correct answer is 8 .
A capacitor $P$ with capacitance $10 \times 10^{-6} \mathrm{~F}$ is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \mathrm{~F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} C$ (assume capacitor $Q$ does not have any charge initially), the value of $\alpha$ is $\_\_\_\_$ .
Explanation:
When capacitor P is fully charged by the 6.0 V battery, it stores an initial amount of charge ( $\mathrm{Q}_{\text {initial }}$ ).
$ Q=C \cdot V $
Given the values for capacitor $P$ the capacitance is $C_P=10 \times 10^{-6} \mathrm{~F}$
$ Q_{\text {initial }}=\left(10 \times 10^{-6}\right) \times 6.0 \mathrm{C} $
$\Rightarrow $ $ Q_{\text {initial }}=60 \times 10^{-6} \mathrm{C} $
After being disconnected from the battery, capacitor $P$ is connected across an initially uncharged capacitor $Q$.
The charges flow from P to Q until both capacitors reach the same electrical potential.
According to the law of conservation of charge, the total charge in the system remains constant :
$ Q_{\text {total }}=Q_{\text {initial }}=60 \times 10^{-6} \mathrm{C} $
In a parallel connection, the equivalent capacitance ( $\mathrm{C}_{\mathrm{eq}}$ ) is the sum of the individual capacitances :
$ C_{\mathrm{eq}}=C_{\mathrm{P}}+C_{\mathrm{Q}} $
$\Rightarrow $ $C_{\mathrm{eq}}=\left(10 \times 10^{-6}\right)+\left(20 \times 10^{-6}\right)$
$\Rightarrow $ $C_{\mathrm{eq}}=30 \times 10^{-6} \mathrm{~F}$
$\Rightarrow $ $\mathrm{V}_{\text {common }}=\frac{\mathrm{Q}_{\text {total }}}{\mathrm{C}_{\mathrm{eq}}}$
$\Rightarrow $ $ \mathrm{V}_{\text {common }}=\frac{60 \times 10^{-6}}{30 \times 10^{-6}} \mathrm{~V}=2.0 \mathrm{~V} $
The final charge $\left(\mathrm{Q}_{\mathrm{Q}}\right)$ using the standard capacitor formula :
$ \mathrm{Q}_{\mathrm{Q}}=\mathrm{C}_{\mathrm{Q}} \cdot \mathrm{~V}_{\text {common }} $
$\Rightarrow $ $\mathrm{Q}_{\mathrm{Q}}=\left(20 \times 10^{-6}\right) \times 2.0 \mathrm{C}$
$\Rightarrow $ $ \mathrm{Q}_{\mathrm{Q}}=40 \times 10^{-6} \mathrm{C} $
So, the charge on capacitor Q will be $4 \times 10^{-5} \mathrm{C}$.
$ \alpha \times 10^{-5} \mathrm{C}=4.0 \times 10^{-5} \mathrm{C} \Rightarrow \alpha=4 $
Space between the plates of a parallel plate capacitor of plate area 4 cm2 and separation of 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is _ pF.
(Given $ \epsilon_0 = 8.85 \times 10^{-12} $ F/m)
Explanation:
$\begin{aligned} & \mathrm{C}_1=\frac{5 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}}=20 \mathrm{pF} \\ & \mathrm{C}_2=\frac{3 \times 4 \times 10^{-4} \times 8.85 \times 10^{-12}}{\frac{1.77}{2} \times 10^{-3}}=12 \mathrm{pF} \\ & \mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}_1 \mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}=\frac{12 \times 20}{12+20}=7.5 \mathrm{pF} \end{aligned}$
Finally equivalent capacitance
$\left(\mathrm{C}_{\text {eq }}\right)_{\text {final }}=7.5+7.5=15 \mathrm{pF}$
Explanation:
To find the dielectric constant $ K $ of the slab, we use the formula for the induced charge $ Q_{\text{in}} $ in a parallel plate capacitor:
$ Q_{\text{in}} = Q \left(1 - \frac{1}{K}\right) $
Given:
Total charge $ Q = 5 \times 10^{-6} \, \text{C} $
Induced charge $ Q_{\text{in}} = 4 \times 10^{-6} \, \text{C} $
We substitute the values into the formula:
$ 4 \times 10^{-6} = 5 \times 10^{-6} \left(1 - \frac{1}{K}\right) $
To solve for the dielectric constant $ K $, simplify the equation:
Divide both sides by $ 5 \times 10^{-6} $:
$ \frac{4}{5} = 1 - \frac{1}{K} $
Rearrange to solve for $ \frac{1}{K} $:
$ 1 - \frac{4}{5} = \frac{1}{K} $
$ \frac{1}{5} = \frac{1}{K} $
Invert both sides to find $ K $:
$ K = 5 $
Therefore, the dielectric constant of the slab is $ K = 5 $.
Four capacitors each of capacitance $16 \mu F$ are connected as shown in the figure. The capacitance between points $A$ and $B$ is : _________ (in $\mu F$).
Explanation:
A parallel plate capacitor consisting of two circular plates of radius 10 cm is being charged by a constant current of 0.15 A . If the rate of change of potential difference between the plates is $7 \times 10^8 \mathrm{~V} / \mathrm{s}$ then the integer value of the distance between the parallel plates is
$\left(\right.$ Take, $\left.\epsilon_0=9 \times 10^{-12} \frac{\mathrm{~F}}{\mathrm{~m}}, \pi=\frac{22}{7}\right)$ ____________ $\mu \mathrm{m}$.Explanation:
Given, $r = 10\,cm = {1 \over {10}}\,m$
$I = 0.15\,A$, and ${{dv} \over {dt}} = 7 \times {10^8}v/s$
We know, for a parallel plate capacitor,
$c = {{{\varepsilon _0}A} \over d} \Rightarrow c = {{{\varepsilon _0}\pi {r^2}} \over d}$ .... (1)
In a capacitor,
$Q = CV$
$ \Rightarrow V = {Q \over C}$
by differentiating w.r.t. t
$ \Rightarrow {{dv} \over {dt}} = {1 \over C}{{dQ} \over {dt}}$
$ \Rightarrow {{dv} \over {dt}} = {d \over {{\varepsilon _0}\pi {r^2}}}I$ (As $I = {{dQ} \over {dt}}$)
$ \Rightarrow d = {{{\varepsilon _0}\pi {r^2}} \over I}{{dv} \over {dt}}$ .... (From (1))
$ \Rightarrow d = {{9 \times {{10}^{ - 12}}} \over {0.15}} \times {{22} \over 7} \times {1 \over {100}} \times 7 \times 10$
$ = {{198} \over {0.15}} \times {10^{ - 6}}$
$ = 1320 \times {10^{ - 6}}$ m
$ \Rightarrow d = 1320\,\mu m$
At steady state the charge on the capacitor, as shown in the circuit below, is _________ $\mu$C.
Explanation:
$\begin{aligned} & \mathrm{i}=\left(\frac{5}{25}\right) \\\\ & \mathrm{Q}=\mathrm{CV} \\\\ & \mathrm{Q}=\left(8 \times 10^{-6}\right)\left(\frac{5}{25} \times 10\right) \\\\ & \mathrm{Q}=\left(\frac{8 \times 5 \times 10^{-2}}{25}\right)=16 \mu \mathrm{C}\end{aligned}$
A capacitor of $10 \mu \mathrm{F}$ capacitance whose plates are separated by $10 \mathrm{~mm}$ through air and each plate has area $4 \mathrm{~cm}^2$ is now filled equally with two dielectric media of $K_1=2, K_2=3$ respectively as shown in figure. If new force between the plates is $8 \mathrm{~N}$. The supply voltage is ________ V.
Explanation:
$\begin{array}{ll} C_1=10 \mu \mathrm{F} & C_2=15 \mu \mathrm{F} \\ F=F_1+F_2 & \end{array}$
$\begin{aligned} & F=\frac{\left(10 \times 10^{-6} V\right)^2}{2 \cdot \frac{A}{2} \cdot \epsilon_0}+\frac{(15~~10)}{2 \cdot \frac{A}{2} \cdot \epsilon_0}=8 \\ & V^2=0.9 \times 10^{-4} \\ & V=0.95 \times 10^{-2} \mathrm{~V} \\ \end{aligned}$
Data in consistent. Answer not matching.
The electric field between the two parallel plates of a capacitor of $1.5 \mu \mathrm{F}$ capacitance drops to one third of its initial value in $6.6 \mu \mathrm{s}$ when the plates are connected by a thin wire. The resistance of this wire is ________ $\Omega$. (Given, $\log 3=1.1$)
Explanation:
To find the resistance of the wire connecting the two plates of the capacitor, we need to apply the formula that relates the time constant $\tau$ (in seconds) of a capacitor-resistor (CR) circuit to the capacitance C (in Farads) and resistance R (in Ohms). The time constant $\tau$ is given by:
$\tau = R \times C$
The time constant also defines the time it takes for the voltage across the capacitor (and therefore the electric field between the plates, since they are directly related) to drop to approximately $\frac{1}{e}$ (where $e$ is the base of natural logarithms, approximately equal to 2.718) of its initial value. However, the question states that the electric field drops to one-third of its initial value. Using the natural logarithm properties, we can relate this decay process to the concept of the time constant.
The formula for the voltage (or electric field) across a discharging capacitor as a function of time $t$ is:
$V(t) = V_0 \times e^{-\frac{t}{RC}}$
where:
- $V(t)$ is the voltage across the capacitor at time $t$,
- $V_0$ is the initial voltage,
- $R$ is the resistance,
- $C$ is the capacitance, and
- $t$ is the time.
Given that the electric field drops to one-third of its initial value in $6.6 \mu \mathrm{s}$, we can set $V(t)$ to $\frac{1}{3}V_0$ and solve for $R$:
$\frac{1}{3}V_0 = V_0 \times e^{-\frac{6.6 \mu s}{R \times 1.5 \mu F}}$
By dividing both sides by $V_0$, we simplify to:
$\frac{1}{3} = e^{-\frac{6.6}{R \times 1.5}}$
Taking the natural logarithm of both sides to solve for $R$:
$\ln\left(\frac{1}{3}\right) = -\frac{6.6}{R \times 1.5}$
Given that $\ln\left(\frac{1}{3}\right) = \ln(3^{-1}) = -\ln(3) = -1.1$ (since $\log 3 = 1.1$ and using the natural log instead of common log), we have:
$-1.1 = -\frac{6.6}{R \times 1.5}$
Solving for $R$:
$R = \frac{6.6}{1.5 \times 1.1}$
Now, calculate the value of $R$:
$R = \frac{6.6}{1.65} = 4\, \Omega$
Therefore, the resistance of the wire connecting the two plates of the capacitor is $4 \, \Omega$.
Three capacitors of capacitances $25 \mu \mathrm{F}, 30 \mu \mathrm{F}$ and $45 \mu \mathrm{F}$ are connected in parallel to a supply of $100 \mathrm{~V}$. Energy stored in the above combination is E. When these capacitors are connected in series to the same supply, the stored energy is $\frac{9}{x} \mathrm{E}$. The value of $x$ is _________.
Explanation:
$E=\frac{1}{2}(25+30+45)(100)^2 \quad \text{.... (i)}$
$\text { Also, } \frac{9}{x} E=\frac{1}{2} \frac{1}{\left(\frac{1}{25}+\frac{1}{30}+\frac{1}{45}\right)}(100)^2 \quad \text{.... (ii)}$
From (i) and (ii)
$x=86$
A parallel plate capacitor of capacitance $12.5 \mathrm{~pF}$ is charged by a battery connected between its plates to potential difference of $12.0 \mathrm{~V}$. The battery is now disconnected and a dielectric slab $(\epsilon_{\mathrm{r}}=6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ $\times10^{-12} \mathrm{~J}$.
Explanation:
$\begin{aligned} E_1 & =\frac{1}{2}\left(\frac{25}{2}\right) \times 10^{-12} \times 144 \\ & =900 \times 10^{-12} \mathrm{~J} \\ E_2 & =\frac{1}{2}\left(6 \times \frac{25}{2} \times 10^{-12}\right)\left(\frac{12}{6}\right)^2=150 \times 10^{-12} \mathrm{~J} \\ \Delta E & =750 \times 10^{-12} \mathrm{~J} \end{aligned}$
Explanation:
In steady state there will be no current in branch of capacitor, so no voltage drop across $\mathrm{R}_2=5 \Omega$
$ \begin{aligned} & \mathrm{I}_2=0 \\\\ & I_1=I_3=\frac{10}{4+6}=1 \mathrm{~A} \\\\ & \mathrm{~V}_{\mathrm{R}_3}=\mathrm{V}_{\mathrm{c}}+\mathrm{V}_{\mathrm{R}_2}, \quad \mathrm{~V}_{\mathrm{R}_2}=0 \end{aligned} $
$\begin{aligned} & \mathrm{I}_3 \mathrm{R}_3=\mathrm{V}_{\mathrm{c}} \\\\ & \mathrm{V}_{\mathrm{c}}=1 \times 6=6 \text { volt } \\\\ & \mathrm{q}_{\mathrm{c}}=\mathrm{CV}_{\mathrm{c}}=10 \times 6=60 \mu \mathrm{C}\end{aligned}$
A parallel plate capacitor with plate separation $5 \mathrm{~mm}$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2 \mathrm{~mm}$, while keeping the battery connections intact, the capacitor draws $25 \%$ more charge from the battery than before. The dielectric constant of the sheet is _________.
Explanation:
Without dielectric
$\mathrm{Q}=\frac{\mathrm{A} \in_0}{\mathrm{~d}} \mathrm{~V}$
with dielectric
$Q=\frac{A \in_0 V}{d-t+\frac{t}{K}}$
given
$\begin{aligned} & \frac{\mathrm{A} \in_0 \mathrm{~V}}{\mathrm{~d}-\mathrm{t}+\frac{\mathrm{t}}{\mathrm{K}}}=(1.25) \frac{\mathrm{A} \in_0 \mathrm{~V}}{\mathrm{~d}} \\ & \Rightarrow 1.25\left(3+\frac{2}{\mathrm{~K}}\right)=5 \\ & \Rightarrow \mathrm{K}=2 \end{aligned}$
A capacitor of capacitance $\mathrm{C}$ and potential $\mathrm{V}$ has energy $\mathrm{E}$. It is connected to another capacitor of capacitance $2 \mathrm{C}$ and potential $2 \mathrm{~V}$. Then the loss of energy is $\frac{x}{3} \mathrm{E}$, where $x$ is _______.
Explanation:
$\begin{aligned} & \text { Energy loss }=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2 \\ & =\frac{2}{3} \cdot E \\ & \therefore x=2 \end{aligned}$
In the given figure, the charge stored in $6 \mu F$ capacitor, when points $A$ and $B$ are joined by a connecting wire is __________ $\mu C$.
Explanation:
At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.
$\begin{aligned} & \mathrm{R}_{\mathrm{eq}}=9 \Omega \\ & \mathrm{i}=\frac{9 \mathrm{~V}}{9 \Omega}=1 \mathrm{~A} \\ & \Delta \mathrm{V}_{6 \Omega}=1 \times 6=6 \mathrm{~V} \\ & \mathrm{~V}_{\mathrm{A}}=3 \mathrm{~V} \end{aligned}$
So, potential difference across 6$\mu$F is 6 V.
$\begin{aligned} \text { Hence } \mathrm{Q} & =\mathrm{C} \Delta \mathrm{V} \\ & =6 \times 6 \times 10^{-6} \mathrm{C} \\ & =36 \mu \mathrm{C} \end{aligned}$
A $16 \Omega$ wire is bend to form a square loop. A $9 \mathrm{~V}$ battery with internal resistance $1 \Omega$ is connected across one of its sides. If a $4 \mu F$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\frac{x}{2} \mu J$, where $x=$ _________
Explanation:
$\begin{aligned} & I=\frac{V}{R_{\text {eq }}} I=\frac{V}{R_{\text {eq }}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4} \\ & I_1=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16} \\ & V_A-V_B=I_1 \times 8=\frac{9}{16} \times 8=\frac{9}{2} V \\ & \therefore U=\frac{1}{2} \times 4 \times \frac{81}{4} \mu J \\ & \therefore U=\frac{81}{2} \mu J \\ & \therefore x=81 \end{aligned}$
The charge accumulated on the capacitor connected in the following circuit is _______ $\mu \mathrm{C}$ (Given $\mathrm{C}=150 \mu \mathrm{F})$
Explanation:
$\begin{aligned} & \mathrm{V}_{\mathrm{A}}+\frac{10}{3}(1)-6(1)=\mathrm{V}_{\mathrm{B}} \\ & \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-\frac{10}{3}=\frac{8}{3} \mathrm{volt} \\ & \mathrm{Q}=\mathrm{C}\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}\right) \\ & =150 \times \frac{8}{3}=400 \mu \mathrm{C} \end{aligned}$
Explanation:
Charge on $\mathrm{C}_2$ is $\mathrm{Q}_2=\mathrm{x} \times 10=10 \mathrm{x} \mu \mathrm{C}$ .....(ii)
Charge on $\mathrm{C}_3$ is $\mathrm{Q}_3=3 \times 10=30 \mu \mathrm{C}$ .....(iii)
Total charge $20+10 \mathrm{x}+30=100$ $\Rightarrow x=5$
In the circuit shown, the energy stored in the capacitor is $n ~\mu \mathrm{J}$. The value of $n$ is __________
Explanation:
$ \begin{aligned} & \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}=3 \mathrm{I}_1=3 \mathrm{~V} \\\\ & \mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}=2 \times 4=8 \mathrm{~V} \end{aligned} $
Subtracting eq. (1) from eq. (2)
$ \begin{aligned} & \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{D}}=5 \mathrm{~V} \Rightarrow \mathrm{V}=5 \mathrm{~V} \\\\ & \mathrm{U}=\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \times 6 \times 5^2=75 \mu \mathrm{J} \end{aligned} $
In the given circuit, $\mathrm{C}_{1}=2 \mu \mathrm{F}, \mathrm{C}_{2}=0.2 \mu \mathrm{F}, \mathrm{C}_{3}=2 \mu \mathrm{F}, \mathrm{C}_{4}=4 \mu \mathrm{F}, \mathrm{C}_{5}=2 \mu \mathrm{F}, \mathrm{C}_{6}=2 \mu \mathrm{F}$, The charge stored on capacitor $\mathrm{C}_{4}$ is ____________ $\mu \mathrm{C}$.
Explanation:
$ \begin{aligned} & \mathrm{C}_{\mathrm{eq}}=0.5 \mu \mathrm{F} \\\\ & \mathrm{Q}=0.5 \times 10=5 \mu \mathrm{C} \\\\ & \mathrm{Q}^{\prime}=\frac{5 \mu \mathrm{C} \times 0.8}{0.8+0.2}=4 \mu \mathrm{C} \end{aligned} $
A $600 ~\mathrm{pF}$ capacitor is charged by $200 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 ~\mathrm{pF}$ capacitor. Electrostatic energy lost in the process is ____________ $\mu \mathrm{J}$
Explanation:
The energy stored in a capacitor can be calculated using the formula:
$ U = \frac{1}{2} C V^2 $
where:
- (U) is the energy,
- (C) is the capacitance,
- (V) is the voltage.
Initially, the energy stored in the first capacitor is:
$ U_{\text{initial}} = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (200 \, \text{V})^2 = 0.012 \, \text{J} = 12 \, \mu\text{J}. $
When the charged capacitor is connected to the uncharged capacitor, the charge will distribute equally between them because they have the same capacitance. Therefore, the final voltage across each capacitor is half of the initial voltage, i.e., 100 V.
The energy in each capacitor after the redistribution is:
$ U_{\text{final each}} = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (100 \, \text{V})^2 = 0.003 \, \text{J} = 3 \, \mu\text{J}. $
As there are two capacitors, the total final energy is:
$ U_{\text{final total}} = 2 \times U_{\text{final each}} = 2 \times 3 \, \mu\text{J} = 6 \, \mu\text{J}. $
The energy loss is the difference between the initial energy and the final energy:
$ \Delta U = U_{\text{initial}} - U_{\text{final total}} = 12 \, \mu\text{J} - 6 \, \mu\text{J} = 6 \, \mu\text{J}. $
As shown in the figure, two parallel plate capacitors having equal plate area of $200 \mathrm{~cm}^{2}$ are joined in such a way that $a \neq b$. The equivalent capacitance of the combination is $x \in_{0} \mathrm{~F}$. The value of $x$ is ____________.
Explanation:
The situation is equivalent to a conducting slab placed between the plates
A parallel plate capacitor with plate area $\mathrm{A}$ and plate separation $\mathrm{d}$ is filled with a dielectric material of dielectric constant $K=4$. The thickness of the dielectric material is $x$, where $x < d$.
Let $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ be the capacitance of the system for $\chi=\frac{1}{3} d$ and $\mathcal{X}=\frac{2 d}{3}$, respectively. If $\mathrm{C}_{1}=2 \mu \mathrm{F}$ the value of $\mathrm{C}_{2}$ is __________ $\mu \mathrm{F}$
Explanation:
$\Rightarrow C_1=\frac{\varepsilon_0 A}{\left(\frac{\frac{d}{3}}{K}+\frac{2 d}{3}\right)}=\frac{\varepsilon_0 A}{\left(\frac{d}{3 \times 4}+\frac{2 d}{3}\right)}$
$ \Rightarrow 2 \mu F=\frac{\varepsilon_0 A \times 12}{9 d}=\frac{4}{3} \frac{\varepsilon_0 A}{d} $
$\Rightarrow 2 \mu F=\frac{4}{3} \frac{\varepsilon_0 A}{d} \quad\left(\right.$ for $\left.x=\frac{1}{3} d\right)$
$\Rightarrow \frac{\varepsilon_0 A}{d}=\frac{3}{2} \mu \mathrm{F}$
Now,
$ \begin{aligned} & C_2=\frac{\varepsilon_0 A}{\left(\frac{\frac{2 d}{3}}{K}+\frac{d}{3}\right)} \quad\left(\text { for } x=\frac{2 d}{3}\right) \\\\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\\\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\\\ & \Rightarrow C_2=\frac{12 \varepsilon_0 A}{2 d+4 d}=\frac{2 \varepsilon_0 A}{d}=\frac{3}{2} \times 2 \\\\ & =3 \mu \mathrm{F} \end{aligned} $
Explanation:
$\Rightarrow i_{A B}=\frac{6}{3}=2 \mathrm{~A} $
$ i_{A D}=\frac{6}{12}=0.5 \mathrm{~A}$
$\Rightarrow V_{B}+2 \times 2-10 \times 0.5=V_{D}$
$\Rightarrow V_{B}-V_{D}=1$ volt
(Assuming Dielectric constant $=10$ )
Explanation:
And charge on $\mathrm{C}_{2}=\mathrm{CV}$
When they are connected in parallel charge will be equally divided so charge on one capacitor is
$q=\frac{K+1}{2} \mathrm{CV}$
So, $V=\frac{q}{K C}=\frac{K+1}{2 K}=55 \mathrm{~V}$
A capacitor of capacitance $900 \mu \mathrm{F}$ is charged by a $100 \mathrm{~V}$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $x \times 10^{-} { }^{2} \mathrm{~J}$. The value of $x$ is _____________.
Explanation:
${U_i} = {1 \over 2}C{V^2} = {1 \over 2} \times 900 \times {10^{ - 6}} \times {100^2} = 4.5$ J
As the other capacitor is identical therefore charge is equally divided and potential difference across the capacitors becomes half. So
${U_f} = {1 \over 2}2C{\left( {{V \over 2}} \right)^2} = {1 \over 2} \times 2 \times 900 \times {10^{ - 6}}{\left( {{{100} \over 2}} \right)^2}$
$ = {9 \over 4}$ J = 2.25 J
So, loss in energy $\Delta {U_{loss}} = {U_i} - {U_f}$
= 2.25 J
= 225 $\times$ 10$^{-2}$ J
A capacitor has capacitance 5$\mu$F when it's parallel plates are separated by air medium of thickness d. A slab of material of dielectric constant 1.5 having area equal to that of plates but thickness $\frac{d}{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be __________ $\mu$F.
Explanation:
When completely air filled
$C=5 \mu \mathrm{F}=\frac{\varepsilon_{0} A}{d} \quad...(1)$
When half filled with $K=1.5$
$ \begin{gathered} \frac{1}{C_{\mathrm{eq}}}=\frac{\frac{d}{2}}{\varepsilon_{0} A}+\frac{\frac{d}{2}}{\varepsilon_{0} A K} \\\\ C_{\text {eq }}=\left(\frac{2 K}{K+1}\right) \frac{\varepsilon_{0} A}{d} \quad...(2) \end{gathered} $
From (1) & (2)
$C_{\mathrm{eq}}=\left(\frac{2 \times 1.5}{1.5+1}\right) 5 \mu \mathrm{F}=6 \mu \mathrm{F}$
A parallel plate capacitor with air between the plate has a capacitance of 15pF. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant 3.5. Then the capacitance becomes $\frac{x}{4}$ pF. The value of $x$ is ____________.
Explanation:
Initially
$ \frac{\varepsilon_{0} A}{d}=15 \times 10^{-12} \mathrm{~F} $
Finally
$ \begin{aligned} & \frac{3.5 \varepsilon_{0} A}{2 d}=\frac{x}{4} \times 10^{-12} \mathrm{~F} \\\\ & \therefore \frac{3.5}{2} \times 15=\frac{x}{4} \\\\ & \Rightarrow x=\frac{3.5 \times 15 \times 4}{2}=105 \end{aligned} $As show in the figure, in steady state, the charge stored in the capacitor is ____________ $\times\, 10^{-6}$ C.
Explanation:
At steady state potential difference across capacitor
${V_c} = {{10 \times 100} \over {110}}\,V$
$Q = C{V_c}$
$ = {{1.1 \times {{10}^{ - 6}} \times 10 \times 100} \over {110}}\,C = 10\,\mu C$
A parallel plate capacitor with width $4 \mathrm{~cm}$, length $8 \mathrm{~cm}$ and separation between the plates of $4 \mathrm{~mm}$ is connected to a battery of $20 \mathrm{~V}$. A dielectric slab of dielectric constant 5 having length $1 \mathrm{~cm}$, width $4 \mathrm{~cm}$ and thickness $4 \mathrm{~mm}$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be ____________ $\epsilon_{0}$ J. (Where $\epsilon_{0}$ is the permittivity of free space)
Explanation:
${d_1} = 4 \times {10^{ - 3}}$
${A_1} = 8 \times 4 \times {10^{ - 4}}\,{m^2}$
$V = 20\,V$
${d_2} = 4 \times {10^{ - 3}},$
${A_2} = 4 \times 1 \times {10^{ - 4}}\,{m^2}$
${C_{eq}} = {{({A_1} + 5{A_2} - {A_2}){\varepsilon _0}} \over d} = {{3(16) \times {{10}^{ - 4}}} \over {4 \times {{10}^{ - 3}}}}{\varepsilon _0}$
$\varepsilon = {1 \over 2}{C_{eq}}{V^2} = {3 \over 2}\left( {{4 \over {10}}} \right)(400){\varepsilon _0} = 240{\varepsilon _0}$
A composite parallel plate capacitor is made up of two different dielectric materials with different thickness $\left(t_{1}\right.$ and $\left.t_{2}\right)$ as shown in figure. The two different dielectric materials are separated by a conducting foil $\mathrm{F}$. The voltage of the conducting foil is V.
Explanation:
${{{C_1}} \over {{C_2}}} = {{3 \times {t_2}} \over {{t_1} \times 4}} = {3 \over 2}$
${q \over {{C_1}}} = {v_1},\,{q \over {{C_2}}} = {v_2}$
${{{v_1}} \over {{v_2}}} = {{{C_2}} \over {{C_1}}} = {2 \over 3}$
Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be ___________ V.
Explanation:
${V_{common}} = {{18CV + 54CV} \over {3C + 9C}} = 6\,V$
A capacitor C1 of capacitance 5 $\mu$F is charged to a potential of 30 V using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor C2 of capacitance 10 $\mu$F as shown in figure. When the switch is closed charge flows between the capacitors. At equilibrium, the charge on the capacitor C2 is __________ $\mu$C.
Explanation:
Let the charge q is flown in the circuit.
So using Kirchhoff's law,
${q \over {10}} = {{150 - q} \over 5}$
$q = 100\,\mu C$
A parallel plate capacitor is made up of stair like structure with a plate area A of each stair and that is connected with a wire of length b, as shown in the figure. The capacitance of the arrangement is ${x \over {15}}{{{ \in _0}A} \over b}$. The value of x is ____________.
Explanation:
The circuit is equivalent to 3 capacitors in parallel as shown
${C_{eq}} = {{{\varepsilon _0}A} \over b}\left( {1 + {1 \over 3} + {1 \over 5}} \right) = {{23} \over {15}}{{{\varepsilon _0}A} \over b}$
$ \Rightarrow x = 23$
A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ___________ nJ.
Explanation:
Electrical energy lost $ = {1 \over 2}\left( {{1 \over 2}C{V^2}} \right)$
$ = {1 \over 2} \times {1 \over 2} \times 50 \times {10^{ - 12}} \times {(100)^2}$
$ = {{500} \over 4}$ nJ
$ = 125$ nJ
The equivalent capacitance between points A and B in below shown figure will be __________ $\mu$F.
Explanation:
${C_{eq}} = {{(3 \times 8) \times 8} \over {(3 \times 8) + 8}}$
$ = {{24 \times 8} \over {32}}$
$ = 6\,\mu F$
Explanation:
$\Delta U = {1 \over 2}(KC - C){V^2}$
$\Delta U = {1 \over 2}(2 - 1)C{V^2}$
$\Delta U = {1 \over 2} \times 200 \times {10^{ - 6}} \times 200 \times 200$
$\Delta U = 4$ J
Explanation:
Potential Difference across each resistor = 2V
q = CV
= 50 $\times$ 10$-$6 $\times$ 2 = 100 $\times$ 10$-$6 = 100 $\mu$C
Explanation:
At time t
$q = Q{e^{ - t/RC}}$
$I =$ ${{dq} \over {dt}}$ = ${Q \over {RC}}{e^{ - t/RC}}$
at $t = 0,I = {Q \over {RC}}{e^{ - 0/RC}}$
$I = {Q \over {RC}} = {{30} \over {5 \times {{10}^6} \times 3}} = 2 \times {10^{ - 6}}$ A
$ \Rightarrow $ I = 2$\mu$A
The value of $\varepsilon $0 = 8.85 $\times$ 10$-$12 F.m$-$1.
The value of 'x' to the nearest integer is _____________.
Explanation:
Separation (d) = 10 m
Thickness = 5 m
Dielectric constant (K) = 10
${c_1} = {{KA{\varepsilon _0}} \over d},{c_2} = {{A{\varepsilon _0}} \over d}$
$ \Rightarrow $ ${c_{eq}} = {{{c_1}{c_2}} \over {{c_1} + {c_2}}} = {{{{KA{\varepsilon _0}} \over d} \times {{A{\varepsilon _0}} \over d}} \over {{{KA{\varepsilon _0}} \over d} + {{A{\varepsilon _0}} \over d}}}$
$ \Rightarrow $ ${c_{eq}} = {{K{A^2}{\varepsilon _0}^2} \over {{d^2}}} \times {d \over {A{\varepsilon _0}(1 + K)}}$
$ \Rightarrow $ ${c_{eq}} = {{KA{\varepsilon _0}} \over {d(1 + K)}} = {{10 \times 100 \times 8.85 \times {{10}^{ - 12}}} \over {5(1 + 10)}}$
$ \Rightarrow $ ${c_{eq}} = {{8.85 \times {{10}^{ - 9}}} \over {55}} = 0.1609090 \times {10^{ - 9}}$
$ \Rightarrow $ ${C_{eq}} = 160.90 \times {10^{ - 12}}$
$ \Rightarrow $ ${C_{eq}} = 161$ PF
Explanation:
When battery is removed & the capacitor is connected
2V + 8v = 20
10V = 20
V = 2 volt
$ \because $ Q = CV
Q = 8 $\times$ 2 = 16$\mu$c
Explanation:
$ = {1 \over 2} \times 14 \times {(12)^2}$ pJ
= 1008 pJ
${U_f} = {{{Q^2}} \over {2kC}}$
$ = {{{{(14 \times 12)}^2}} \over {2 \times 7 \times 14}}$
= 144 pJ
oscillating energy = Ui $-$ Uf
= 1008 $-$ 144
= 864 pJ
Explanation:
${C_{eq}} = {{2C \times C} \over {2C + C}} = {{2C} \over 3} $
$= {2 \over 3}{{{ \in _0}A} \over d} = {2 \over 3} \times {{{ \in _0}} \over d} \times \left( {2 \times {3 \over 2}} \right) = {{2{\varepsilon _0}} \over d}$
Explanation:
${1 \over C} = {1 \over {{C_1}}} + {1 \over {{C_2}}} = {d \over {2K{\varepsilon _0}A}} + {d \over {2{\varepsilon _0}A}}$
${1 \over C} = {d \over {2{\varepsilon _0}A}}\left( {{{K + 1} \over K}} \right)$
$C = {{2{\varepsilon _0}AK} \over {d(K + 1)}} = {{2 \times 2 \times 3.2} \over {1 \times 4.2}}{\varepsilon _0} = 3.04{\varepsilon _0}$
Explanation:
Capacitor is fully charged
So no current is there in branch ADB
$ \therefore $ Effective circuit of current flow :
Req = $\left( {{{4 \times 2} \over {4 + 2}}} \right) + 2$
Req = ${4 \over 3} + 2$ = ${{10} \over 3}\Omega $
i = ${{10} \over {{{10} \over 3}}}$ = 3A
So potential different across AEB
= 2 × 1 + 2 × 3 = 8V
Hence potential difference across Capacitor = 8V
Explanation:
Final common potential
= $\frac{220\times 5+0\times 2.5}{5+2.5} $ = 220 $ \times $ $\frac{2}{3} $
uf = $\frac{1}{2} $ $ \times $ (5 + 2.5)$ \times $10-6 $ \times $ $\left( 220\times \frac{2}{3} \right)^{2} $
$\Delta $u = ui - uf
$ \Rightarrow $ $\Delta $u = –403.33 × 10–4
$ \Rightarrow $ –403.33 × 10–4 = ${X \over {100}}J$
$ \Rightarrow $ X = -4.03
Value of X is approximate 4
Explanation:
Ui = ${1 \over 2}CV_0^2$
Uf = ${1 \over 2} \times 2C \times {\left( {{{{V_0}} \over 2}} \right)^2}$
$\Delta $E = ${1 \over 2}CV_0^2$ - ${1 \over 2} \times 2C \times {\left( {{{{V_0}} \over 2}} \right)^2}$
= ${{{CV_0^2} \over 4}}$
= ${1 \over 4} \times 60 \times {10^{ - 12}} \times 4 \times {10^2}$
= 6 $ \times $ 10-9 = 6 nJ
The magnitude of q1 is ________________.
Explanation:
Switch connected to position P
${V_A} - 1\,.\,{i_1} - 1 + 2 - 2{i_1} = {V_A}$
$3{i_1} = 1$
${i_1} = {1 \over 3}A$
Now, ${V_A} - 1\,.\,{i_1} - 1 = {V_B}$
${V_A} - {V_B} = 1 + {i_1} = {4 \over 3}V$
Potential drop across capacitor, $\Delta V = {4 \over 3}V$
$\therefore$ Charge on capacitor, ${q_1} = C\Delta V = 1 \times {4 \over 3}\mu C$
${q_1} = 1.33\mu C$
The magnitude of q2 is ________________.
Explanation:
${V_A} - 1\,.\,{i_2} + 2 - 2{i_2} = {V_A}$
$3{i_2} = 2$
$ \Rightarrow {i_2} = {2 \over 3}A$
Now, ${V_A} - \,{i_2} \times 1 = {V_B}$
${V_A} - {V_B} = {i_2} \times 1 = {2 \over 3}V$
Potential difference across capacitor, $\Delta V = {2 \over 3}V$
$\therefore$ Charge on capacitor, ${q_2} = C\Delta V = 1 \times {2 \over 3} = 0.67\mu C$
Explanation:
Combine capacitance of series combination
$ \begin{aligned} \frac{1}{C} & =\frac{1}{C_1}+\frac{1}{C_2} ~~~~....(1)\\\\ \Rightarrow & \frac{1}{C}=\frac{1}{2000}+\frac{1}{3000}=\frac{(6+4)}{12000} \\\\ \frac{1}{C} & =\frac{10}{12000} \Rightarrow C=1200 \mathrm{pF} \end{aligned} $
Differentiating Eq. (1), we get
$ \begin{aligned} -\frac{1}{C^2} \cdot d C & =-\frac{1}{C_1^2} d C_1+\left(-\frac{1}{C_2^2}\right) \cdot d C_2 \\\\ \Rightarrow \frac{d C}{C^2} & =\left(\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right) \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{d C_1}{C_1^2}+\frac{d C_2}{C_2^2}\right] \cdot C \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{10}{(2000)^2}+\frac{15}{(3000)^2}\right] \times 1200 \\\\ \Rightarrow \frac{d C}{C} & =\left[\frac{10}{4} \times 10^{-6}+\frac{15}{9} \times 10^{-6}\right] \times 1200 \\\\ & =\left[\frac{10}{4}+\frac{15}{9}\right] \times 12 \times 10^{-4} \end{aligned} $
$ \Rightarrow \frac{d C}{C}=[2.5+1.66] \times 12 \times 10^{-4}=[4.17 \times 12] \times 10^{-4} $
And $ \frac{d V}{V}=\frac{0.02}{5} $
Energy stored in the capacitor is given by
$ U=\frac{1}{2} C V^2 $ .....(2)
Differentiating Eq. (2), we get
$ \begin{aligned} \frac{d U}{U} \times 100 & =\frac{d C}{C} \times 100+2 \frac{d V}{V} \times 100 \\\\ & =\left[50.04 \times 10^{-4} \times 100+\frac{2 \times 0.02}{5} \times 100\right] \\\\ \left(\frac{d U}{U}\right)^{100} & =[0.5004+0.8] \% \Rightarrow 1.30 \% \end{aligned} $