SATHEE: electrostatic-potential-and-capacitance Question 29

इलेक्ट्रोस्टैटिक क्षमता और धारिता

Question: Q. 4. A slab of material of dielectric constant $κ$ has the same area as that of the plates of a parallel plate capacitor but has the thickness $d / 2$ , where $d$ is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

U [O.D. I, II, III 2013]

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Solution:

Ans. Initially when there is vacuum between the two plates, the capacitance of the two parallel plates is,

$C_{0} = \frac{ε_{0} A}{d}$

where, $A$ is the area of parallel plates.

Suppose that the capacitor is connected to a battery, an electric field $E_{0}$ is produced.

Now if we insert the dielectric slab of thickness $t = d / 2$ , the electric field reduces to $E$ .

Now the gap between the plates is divided in two parts, for distance $t$ there is electric field $E$ and for the remaining distance $(d - t)$ , the electric field is $E_{0}$ . $1 / 2$ If $V$ be the potential difference between the plates of the capacitor, then

$\begin{aligned} V & = E t + E_{0} (d - t) V & = \frac{E d}{2} + \frac{E_{0} d}{2} & = \frac{d}{2} (E + E_{0}) \Rightarrow V & = \frac{d}{2} (\frac{E_{0}}{κ} + E_{0}) & = \frac{d E_{0}}{2 κ} (κ + 1) \end{aligned}$

$\begin{aligned} V & = \frac{E d}{2} + \frac{E_{0} d}{2} (∵ t = \frac{d}{2}) & = \frac{d}{2} (E + E_{0}) (∵ \frac{E_{0}}{E} = κ) \Rightarrow V & = \frac{d}{2} (\frac{E_{0}}{κ} + E_{0}) & = \frac{d E_{0}}{2 κ} (κ + 1) Now, E_{0} & = \frac{σ}{ε_{0}} = \frac{q}{ε_{0} A} \Rightarrow V & = \frac{d}{2 κ} \cdot \frac{q}{ε_{0} A} (κ + 1) \end{aligned}$

We know,

$C = \frac{q}{V} = \frac{2 κ ε_{0} A}{d (κ + 1)}$

इलेक्ट्रोस्टैटिक क्षमता और धारिता

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इलेक्ट्रोस्टैटिक क्षमता और धारिता

विषयसूची

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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