SATHEE: Electrostatics 5 Question 14

Electrostatics 5 Question 14

15. An electric line of force in the $x - y$ plane is given by the equation $x^{2} + y^{2} = 1$ . A particle with unit positive charge, initially at rest at the point $x = 1, y = 0$ in the $x - y$ plane, will move along the circular line of force.

$(1988, 2 M)$

Analytical & Descriptive Questions

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Solution:

In the given arrangement, capacitor can be viewed as threedifferent capacitors connected in parallel as shown below,

So, equivalent capacity of the system is

$\begin{aligned} C_{e q} & = C_{1} + C_{2} + C_{3} \\ \Rightarrow \frac{K ε_{0} A}{d} & = \frac{K_{1} ε_{0} A / 3}{d} + \frac{K_{2} ε_{0} A / 3}{d} + \frac{K_{3} ε_{0} A / 3}{d} \\ \Rightarrow K & = \frac{K_{1}}{3} + \frac{K_{2}}{3} + \frac{K_{3}}{3} \\ Here, K_{1} & = 10, K_{2} = 12 and K_{3} = 14 \\ So, K & = \frac{10 + 12 + 14}{3} \Rightarrow K = 12 \end{aligned}$

16 This capacitor system can be converted into two parts as shown in the figure

$d / 2 d / 2$

where $C_{1}, C_{2}, C_{3}$ and $C_{4}$ are capacitance of the capacitor having dielectric constants $K_{1}, K_{2}, K_{3}$ and $K_{4}$ respectively.

Here, $C_{1} = \frac{K_{1} ε_{0} A / 2}{d / 2} = \frac{K_{1} ε_{0} A}{d}$

Similarly, $C_{2} = \frac{K_{2} ε_{0} A}{d}, C_{3} = \frac{K_{3} ε_{0} A}{d}$ and $C_{4} = \frac{K_{4} ε_{0} A}{d}$

Since, equivalent capacitance in series combination is

$C_{e q} = \frac{C_{1} \cdot C_{2}}{C_{1} + C_{2}}$

Here, $C_{1}, C_{2}$ and $C_{3}, C_{4}$ are in series combination.

$\begin{aligned} ∴ {(C_{eq})}_{12} & = \frac{C_{1} \cdot C_{2}}{C_{1} + C_{2}} = \frac{\frac{K_{1} ε_{0} A}{d} \cdot \frac{K_{2} ε_{0} A}{d}}{\frac{K_{1} ε_{0} A}{d} + \frac{K_{2} ε_{0} A}{d}} \\ = \frac{K_{1} \cdot K_{2}}{K_{1} + K_{2}} \cdot \frac{ε_{0} A}{d} \end{aligned}$

Similarly, ${(C_{e q})}_{34} = \frac{K_{3} \cdot K_{4}}{K_{3} + K_{4}} \cdot \frac{ε_{0} A}{d}$

Now, ${(C_{eq})}_{12}$ and ${(C_{eq})}_{34}$ are in parallel combination.

$∴ C_{net} = {(C_{eq})}_{12} + {(C_{eq})}_{34}$

$= \frac{K_{1} \cdot K_{2}}{K_{1} + K_{2}} \cdot \frac{ε_{0} A}{d} + \frac{K_{3} \cdot K_{4}}{K_{3} + K_{4}} \cdot \frac{ε_{0} A}{d}$

$\Rightarrow C_{net} = \frac{K_{1} \cdot K_{2}}{K_{1} + K_{2}} + \frac{K_{3} \cdot K_{4}}{K_{3} + K_{4}} \frac{ε_{0} A}{d}$

If $K$ is effective dielectric constant, then

$C_{net} = \frac{K ε_{0} A}{d}$

From Eqs. (i) and (ii),

$\begin{aligned} \frac{K ε_{0} A}{d} & = \frac{K_{1} \cdot K_{2}}{K_{1} + K_{2}} + \frac{K_{3} \cdot K_{4}}{K_{3} + K_{4}} \frac{ε_{0} A}{d} \\ K & = \frac{K_{1} \cdot K_{2}}{K_{1} + K_{2}} + \frac{K_{3} \cdot K_{4}}{K_{3} + K_{4}} \end{aligned}$