Question

17) Calculate the capacitance of a parallel plate capacitor with dielectric medium 6 if
the separation between the plates 8.85 mm and plate area 6 x 10^-4 m2.
​

Answer 1

Answer:

3.6 PF

Explanation:

★ PARALLEL PLATE CAPACITOR ★

★ GIVEN ;

» Dielectric Medium ( K ) = 6

» Separation betweenplates(d)=8.85 mm

» Area of plates ( A ) = 6 × 10-⁴ m²

★ PERMITTIVITY CONSTANT ( Ë∅ ) = 8.85 × 10-¹² ★

→ CAPACITANCE ( C ) = ????

★» BY USING FORMULA ; ♣

♣ NOTE :- FORMULA ONLY VALID WHEN DIELECTRIC MEDIUM ( K ) IS INSERTED ♥

★» [ C = (Ë∅ • K • A) / d ] «★

» C = ( Ë∅ × 6 × 6 × 10-⁴ ) / 8.85 × 10-³ m

» C = (8.85 × 10-¹² × 36 × 10-⁴)/ 8.85 ×10-³

» C = ( 36 × 10^-16 ) / 10-³

» C = 36 × 10-¹³

★» C = 3.6 PF [ Pico Farad ]

ANSWER :- HENCE ;

Capacitance between plates of Area

Capacitance between plates of Area 6 × 10-⁴ m² is ; " 3.6 PF " .

Answer 2

♊♊YOUR ANSWERS♍♍

✍C = 3.6 PF

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EXPLANATION —

↪Actually welcome to the concept of Capacitance

↪Here we will use general Dielectric constant Capacitance Formula

Given—

✏Dielectric Constant(k) = 6

✏Separation b/w the plates(d)= 8.85 mm= 8.85 × 10¯³ m

✏Area of Each plates (A) = 6 × 10¯⁴ m²

We know that ∈° =8. 85 × 10¯¹²

To find

the Capacitance =?

Now by dielectric formula to find the Capacitance (C)

C = ∈° k A/d

Putting the values

C = (8. 85 × 10¯¹² × 6 ×6 × 10¯⁴) /8.85 × 10¯³

On solving

C = 36 × 10¯¹³ μF

We can also write as

C = 3.6 PF

Here PF is Pico Faraday

Therefore the required capacitor is 3.6 PF

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