Question

The capacitance of a parallel plate air capacitor is 12 pF. If the separation between its plates is doubled, what will be the new capacitance?​

Answer 1

Answer:

o. o

Explanation:

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Answer 2

The capacitance of a parallel plate air capacitor is 12 pF. If the separation between its plates is doubled, what will be the new capacitance?​

Given:

$C=12pF$,

$d^1=d/2$,

$A^1=2A$.

Calculation:

capacity of capacitor,$C=E.2A/d$=$12pF$

New capacity can be given by,

$C^1=E.2A^1/d^1$

$C^1=E.2A/d/2$

$C^1=4E.A/d$

$C^1=4*12$

$C^1=48pF$

Explanation:

• The capacitance of a parallel plate capacitor is $C=E0AdC=E0Ad$, when the plates are separated by air or free space.
• A parallel plate  capacitor with a dielectric between its plates has a capacitance given by $C=kee0AdC=ke0Ad$,
• Where k is the dielectric constant of the material.
• The capacitance of a parallel plate air capacitor is $12uF$.
• The separation between the plates is $8mm$.
• Two dielectric slabs of the same size fill the air space.