
Capacity of a parallel capacitor with dielectric constant $5$ is $40\mu F$.Calculate the capacity of the same capacitor when dielectric material is removed.
Answer
497.1k+ views
Hint:To calculate the capacity of the capacitor, we use the concept of capacitor with and without dielectric and the effect of dielectric on the capacity of the capacitor. We will also discuss dielectric used in capacitors that increases the storage capacity of the capacitor.
Formulae used:
The capacity of parallel plate capacitor with dielectric is given by
${C_d} = k{C_{air}}$
Where, $k$ - dielectric constant and ${C_{air}}$ - capacity of parallel plate capacitor with air as dielectric.
Complete step by step answer:
Parallel Plate Capacitors are formed by an arrangement of electrodes and insulating material or dielectric. A parallel plate capacitor can only store a finite amount of energy before dielectric breakdown occurs. Given, $k = 5$ , ${C_d} = 40\mu F$
Using the given formula,
${C_d} = k{C_{air}}$
$\Rightarrow 40 = 5{C_{air}}$
Solving, we get
$\therefore {C_{air}} = 8\mu F$
Hence, the capacity of the same capacitor when dielectric material is removed i.e. air as dielectric is ${C_{air}} = 8\mu F$.
Note: The capacity of a parallel plate capacitor with dielectric constant $k$ is, dielectric constant times the capacity of a parallel plate capacitor with air as dielectric.Dielectric is an insulating material having properties of charge storage having properties of charge storage which increases the capacity of the capacitor.
Formulae used:
The capacity of parallel plate capacitor with dielectric is given by
${C_d} = k{C_{air}}$
Where, $k$ - dielectric constant and ${C_{air}}$ - capacity of parallel plate capacitor with air as dielectric.
Complete step by step answer:
Parallel Plate Capacitors are formed by an arrangement of electrodes and insulating material or dielectric. A parallel plate capacitor can only store a finite amount of energy before dielectric breakdown occurs. Given, $k = 5$ , ${C_d} = 40\mu F$
Using the given formula,
${C_d} = k{C_{air}}$
$\Rightarrow 40 = 5{C_{air}}$
Solving, we get
$\therefore {C_{air}} = 8\mu F$
Hence, the capacity of the same capacitor when dielectric material is removed i.e. air as dielectric is ${C_{air}} = 8\mu F$.
Note: The capacity of a parallel plate capacitor with dielectric constant $k$ is, dielectric constant times the capacity of a parallel plate capacitor with air as dielectric.Dielectric is an insulating material having properties of charge storage having properties of charge storage which increases the capacity of the capacitor.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Explain sex determination in humans with the help of class 12 biology CBSE

