SATHEE: Electrostatics 5 Question 47

Electrostatics 5 Question 47

49. The capacitance of a parallel plate capacitor with plate area $A$ and separation $d$ , is $C$ . The space between the plates is filled with two wedges of dielectric constants $K_{1}$ and $K_{2}$ respectively (figure). Find the capacitance of the resulting capacitor.

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Answer:

Correct Answer: 49. $C_{R} = \frac{C K_{1} K_{2}}{K_{2} - K_{1}} \ln \frac{K_{2}}{K_{1}} where, C = \frac{ε_{0} A}{d}$

Solution:

Let length and breadth of the capacitor be $l$ and $b$ respectively and $d$ be the distance between the plates as shown in figure. Then, consider a strip at a distance $x$ of width $d x$ .

Now, $Q R = x \tan θ$

and $P Q = d - x \tan θ$

where, $\tan θ = d / l$

Capacitance of $P Q$

$\begin{matrix} C_{1} = \frac{K_{1} ε_{0} (b d x)}{d - x \tan θ} = \frac{K_{1} ε_{0} (b d x)}{d - \frac{x d}{l}} \\ C_{1} = \frac{K_{1} ε_{0} b l d x}{d (l - x)} = \frac{K_{1} ε_{0} A (d x)}{d (l - x)} \end{matrix}$

and $C_{2} =$ capacitance of $Q R = \frac{K_{2} ε_{0} b (d x)}{x \tan θ}$

$C_{2} = \frac{K_{2} ε_{0} A (d x)}{x d} \tan θ = \frac{d}{l}$

Now, $C_{1}$ and $C_{2}$ are in series. Therefore, their resultant capacity $C_{0}$ will be given by

$\frac{1}{C_{0}} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

Then, $\frac{1}{C_{0}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{d (l - x)}{K_{1} ε_{0} A (d x)} + \frac{x \cdot d}{K_{2} ε_{0} A (d x)}$

$\begin{aligned} \frac{1}{C_{0}} & = \frac{d}{ε_{0} A (d x)} \frac{l - x}{K_{1}} + \frac{x}{K_{2}} \\ = \frac{d K_{2} (l - x) + K_{1} x}{ε_{0} A K_{1} K_{2} (d x)} \\ ∴ C_{0} & = \frac{ε_{0} A K_{1} K_{2}}{d K_{2} (l - x) + K_{1} x} d x \\ C_{0} & = \frac{ε_{0} A K_{1} K_{2}}{d K_{2} l + (K_{1} - K_{2}) x} d x \end{aligned}$

Now, the net capacitance of the given parallel plate capacitor is obtained by adding such infinitesimal capacitors placed parallel from $x = 0$ to $x = l$

i.e. $C_{R} = \int_{x = 0}^{x = l} C_{0} = \int_{0}^{l} \frac{ε_{0} A K_{1} K_{2}}{d K_{2} l + (K_{1} - K_{2}) x} d x$

Finally we get $C_{R} = \frac{K_{1} K_{2} ε_{0} A}{(K_{2} - K_{1}) d} \ln \frac{K_{2}}{K_{1}}$

$= \frac{C K_{1} K_{2}}{K_{2} - K_{1}} \ln \frac{K_{2}}{K_{1}} where, C = \frac{ε_{0} A}{d}$

Electrostatics 5 Question 47

49. The capacitance of a parallel plate capacitor with plate area $A$ and separation $d$ , is $C$ . The space between the plates is filled with two wedges of dielectric constants $K_{1}$ and $K_{2}$ respectively (figure). Find the capacitance of the resulting capacitor.

Answer:

Solution:

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Electrostatics 5 Question 47

49. The capacitance of a parallel plate capacitor with plate area A and separation d, is C. The space between the plates is filled with two wedges of dielectric constants K1 and K2 respectively (figure). Find the capacitance of the resulting capacitor.

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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49. The capacitance of a parallel plate capacitor with plate area $A$ and separation $d$ , is $C$ . The space between the plates is filled with two wedges of dielectric constants $K_{1}$ and $K_{2}$ respectively (figure). Find the capacitance of the resulting capacitor.