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Physics · 2024 · 5 marks
CBSE 2024 · Region 4 · Set 1 · Q32
(a) (i) A dielectric slab of dielectric constant ' K ' and thickness ' t ' is inserted between plates of a parallel plate capacitor of plate…
(i)
A dielectric slab of dielectric constant ' K ' and thickness ' t ' is inserted between plates of a parallel plate capacitor of plate separation d and plate area A . Obtain an expression for its capacitance.
(ii)
Two capacitors of different capacitances are connected first ($\displaystyle 1$) in series and then ($\displaystyle 2$) in parallel across a dc source of $\displaystyle 100$ V . If the total energy stored in the combination in the two cases are $\displaystyle 40$ mJ and $\displaystyle 250$ mJ respectively, find the capacitance of the capacitors.
(i)
Using Gauss's law, show that the electric field $\displaystyle \overrightarrow{\mathrm{E}}$ at a point due to a uniformly charged infinite plane sheet is given by $\displaystyle \overrightarrow{\mathrm{E}}=\frac{\sigma}{2 \varepsilon_{0}} \hat{\mathrm{n}}$ where symbols have their usual meanings.
(ii)
Electric field $\displaystyle \overrightarrow{\mathrm{E}}$ in a region is given by $\displaystyle \overrightarrow{\mathrm{E}}=\left(5 x^{2}+2\right) \hat{\mathrm{i}}$ where E is in N/C and $\displaystyle x$ is in meters. A cube of side $\displaystyle 10$ cm is placed in the region as shown in figure.
Calculate ($\displaystyle 1$) the electric flux through the cube, and ($\displaystyle 2$) the net charge enclosed by the cube.
Marking-scheme solution
(i)
Electric field in air between plates:
$\displaystyle E_{0}=\frac{\sigma}{\varepsilon_{0}}$
Electric field inside the dielectric:
$\displaystyle E=\frac{\sigma}{\varepsilon_{0} K}$
Potential difference between the plates:
$\displaystyle V=E_{0}(d-t)+E t$
$\displaystyle V=\frac{\sigma}{\varepsilon_{0}}\left[d-t+\frac{t}{K}\right]$
$\displaystyle V=\frac{q}{A \varepsilon_{0}}\left[d-t+\frac{t}{K}\right]$
Capacitance:
$\displaystyle C=\frac{q}{V}$
$\displaystyle C=\frac{A \varepsilon_{0}}{d-t+\dfrac{t}{K}}$
$\displaystyle C=\frac{A \varepsilon_{0}}{d-t\left(1-\dfrac{1}{K}\right)}$
(ii)
Total energy stored in series combination:
$\displaystyle \frac{1}{2}\left(\frac{C_{1} C_{2}}{C_{1}+C_{2}}\right) V^{2}=40 \times 10^{-3} \mathrm{~J}$...............................($\displaystyle 1$)
Energy stored in parallel combination:
$\displaystyle \frac{1}{2}\left(C_{1}+C_{2}\right) V^{2}=250 \times 10^{-3} \mathrm{~J}$........................................($\displaystyle 2$)
Substituting the value of $\displaystyle V=100 \mathrm{~V}$ in eq ($\displaystyle 1$) and ($\displaystyle 2$), on solving:
$\displaystyle C_{1}=4 \times 10^{-5} \mathrm{~F}$ or $\displaystyle 40 \mu \mathrm{F}$
$\displaystyle C_{2}=1 \times 10^{-5} \mathrm{~F}$ or $\displaystyle 10 \mu \mathrm{F}$
(i)
$\displaystyle \oint \vec{E} \cdot d \vec{s}=\int_{1} \vec{E} \cdot d \vec{s}+\int_{2} \vec{E} \cdot d \vec{s}$
$\displaystyle =2 E A$
From Gauss's law:
$\displaystyle \oint \vec{E} \cdot d \vec{s}=\frac{q}{\varepsilon_{0}}$
$\displaystyle 2 E A=\frac{\sigma A}{\varepsilon_{0}}$
$\displaystyle E=\frac{\sigma}{2 \varepsilon_{0}}$
Vectorially $\displaystyle \vec{E}=\frac{\sigma}{2 \varepsilon_{0}} \hat{n}$
Electric field is normally outward of the sheet.
(ii)(1)
Electric flux through the cube:
$\displaystyle \phi=\phi_{L}+\phi_{R}$
$\displaystyle \phi=\int \vec{E}_{L} \cdot d \vec{s}+\int \vec{E}_{R} \cdot d \vec{s}$
$\displaystyle =-2 \times 100 \times 10^{-4}+\left[5 \times\left(10 \times 10^{-2}\right)^{2}+2\right] \times 100 \times 10^{-4}$
$\displaystyle \phi=5 \times 10^{-4} \mathrm{~Nm}^{2} \mathrm{C}^{-1}$
(2)
$\displaystyle \phi=\frac{q_{e n}}{\varepsilon_{0}}$
$\displaystyle q_{e n}=\phi \cdot \varepsilon_{0}$
$\displaystyle =5 \times 10^{-4} \times 8.85 \times 10^{-12}$
$\displaystyle =4.43 \times 10^{-15} \mathrm{C}$
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