Read the following-Then K e q - -A- K 1- K 2 K 3- K 4B- K 2C-D- K 1 K 2- K 3 K 4- K 1- K 2- K 3- K 4

The correct options are A (K_1 + K_2)(K_3 + K_4)/K_1+K_2+K_3+K_4C_1 = K_1 A/2ε_0 /d/2 = K_1A ε _0/d C_2 = K_2A ε_0/d C_3 = K_3A ε_0/d C_4 = K_4A ε_0/d C_eq ...

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Byju's Answer

Standard XII

Mathematics

Vn Method

Read the foll...

Question

Read the following.

Then $K_{e} q = ?$

\frac{(K_{1} + K_{2}) (K_{3} + K_{4})}{K_{1} + K_{2} + K_{3} + K_{4}}

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\frac{(K_{1} + K_{4}) (K_{2} + K_{3})}{K_{1} + K_{3} + K_{4} + K_{2}}

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\frac{(K_{2} + K_{4}) (K_{1} + K_{3})}{K_{1} + K_{2} + K_{3} + K_{4}}

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\frac{K_{1} K_{2} + K_{3} K_{4}}{K_{1} + K_{2} + K_{3} + K_{4}}

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Solution

The correct option is A $\frac{(K_{1} + K_{2}) (K_{3} + K_{4})}{K_{1} + K_{2} + K_{3} + K_{4}}$
$C_{1} = \frac{K_{1} \frac{A}{2} ε_{0}}{\frac{d}{2}} = \frac{K_{1} A ε_{0}}{d}$

$C_{2} = \frac{K_{2} A ε_{0}}{d}$

$C_{3} = \frac{K_{3} A ε_{0}}{d}$

$C_{4} = \frac{K_{4} A ε_{0}}{d}$

$C_{e q} = \frac{C_{1} C_{3}}{C_{1} + C_{3}} + \frac{C_{2} C_{4}}{C_{2} + C_{4}}$

$C_{e q} = K_{e q} = \frac{A ε_{0}}{d} = [\frac{K_{1} K_{3}}{K_{1} + K_{3}} + \frac{K_{2} K_{4}}{K_{2} + K_{4}}] + \frac{A ε_{0}}{d}$

$K_{e q} = [\frac{K_{1} K_{3}}{K_{1} + K_{3}} + \frac{K_{2} K_{4}}{K_{2} + K_{4}}]$

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