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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance2 Marks
Question
X and Y are two parallel plate capacitors, whose plates have the same area and the same distance between them. There is air between the plates of $X$ and a dielectric material of dielectric constant $k=5$ between the plates of $Y$ (in the figure). (a) Calculate the potential diffe-rence between the plates X and Y . (b) What is the ratio of electrical potential energies stored in X and Y ?
✓

Answer

(a) Capacitance of air capacitor
$ C_1=\frac{\in_0 A}{d} $
and capacitance of Y $C _2=\frac{ KA \in_0}{d}$ or $C _2= KC _1$
If $C_1=C$ then $C_2=K C$
$\Rightarrow \quad C_2=5 C$
$\because \quad K = C$
When two capacitors are connected in series, the charge stored on each is equal, therefore
$q= C _1 V_1= C _2 V_2$ or $q=C V_1=5 C V_2$
or $V_1=5 V_2$
But $V_1+V_2=12$
$\therefore \quad 5 V_2+V_2=12$
$6 V_2=12 \Rightarrow V_2=2$ volt
$\therefore \quad V _1=5 V_2=5 \times 2=10$ volt
(b) Since the energy of the charged capacitor
$U =\frac{1}{2} \frac{q^2}{ C }$
$U \propto \frac{1}{ C }$
$\therefore \quad \frac{ U _1}{ U _2}=\frac{ C _2}{ C _1}=\frac{5 C }{ C }=5$
or $U _1: U _2=5: 1$

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