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Electrostatic Potential and Capacitance — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectrostatic Potential and Capacitance3 Marks
Question
Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of $\varepsilon_{\text{r}} = 4.$
  1. Calculate capacitance of each capacitor if equivalent capacitance of the combination is $4\mu\text{f}.$
  2. Calculate the potential difference between the plates of X and Y.
  3. Estimate the ratio of electrostatic energy stored in X and Y.
✓

Answer

  1. Let $\text{C}_{x} = \text{C}$
$\text{C}_{y} = 4\text{C } \text{as it has a dielectric medium of } \varepsilon_{r} = 4 .$

For series combination of two capacitors

$\frac{1}{\text{C}} =\frac{1}{\text{C}_{x}} + \frac{1}{\text{C}_{y}}$

$\Rightarrow\frac{1}{4\mu\text{F}} = \frac{1}{\text{C}} + \frac{1}{4\text{C}}$

$\frac{1}{4\mu\text{F}} = \frac{5}{4\text{C}}$

$\Rightarrow\text{C} = 5\mu\text{F}$

Hence $\text{C}_{X} = 5\mu\text{F}$

$\text{C}_{y} = 20 \mu\text{F}$
  1. Total charge Q = CV
$ = 4 \mu\text{F}\times15 \text{V} = 60 \mu\text{C}$

$\text{V}_{x} = \frac{\text{Q}}{\text{C}_{x}} = \frac{60\mu\text{C}}{5\mu\text{F}} = 12 \text{V}$

$\text{V}_{y} = \frac{\text{Q}}{\text{C}_{y}} = \frac{60\mu\text{C}}{20\mu\text{F}} = 3 \text{V}$
  1. $\frac{{\text{E}}_{\text{x}}}{\text{E}_{\text{y}}} = \frac{^\frac{\text{Q}^{2}}{2{C}_{\text{X}}}}{\frac{\text{Q}^{2}}{2\text{C}_{Y}}} = \frac{\text{C}_{\text{y}}}{\text{C}_{\text{x}}} = \frac{20}{5} = 4:1$

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