A capacitor having capacitance 1 uF with air, is filled with two dielectrics as shown. How many times capacitance will increase

12

6

8/3

3

Solution

After filling with dielectrics the two capacitors will be in parallel order. As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor

C = $\frac{ε_{o} A}{d}$

where A is area of each plate and d is the separation between the plates.

After filling with dielectrics, we have two capacitors of capacitance.

$C_{1} = \frac{K_{1} ε_{o} (\frac{A}{2})}{d} C_{2} = \frac{8}{2} \frac{ε_{o} A}{d} C_{2} = \frac{4 ε_{o} A}{d} C_{2} = 4 C$

$And C_{2} = \frac{K_{2} ε_{o} (\frac{A}{2})}{d} C_{2} = \frac{4}{2} \frac{ε_{o} A}{d} C_{2} = \frac{2 ε_{o}}{d} C_{2} = 2 C$

Hence, their equivalent capacitance
C_eq = C₁ + C₂

= 4C +2C

C = 6C

ie, new capacitance will be six times of the original.

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Electrostatic Potential And Capacitance

A capacitor having capacitance 1 uF with air, is filled with two dielectrics as shown. How many times capacitance will increase

12

6

8/3

3

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Electrostatic Potential And Capacitance

A capacitor having capacitance 1 uF with air, is filled with two dielectrics as shown. How many times capacitance will increase 12 6 8/3 3

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A capacitor having capacitance 1 uF with air, is filled with two dielectrics as shown. How many times capacitance will increase

12

6

8/3

3