# Capacitance - Short Question Solution | Physics Grade XI

### Capacitance Short Question Solution

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# Capacitance - Short Question Answers

1. Two plates of a charged capacitor are brought closer. What happens to its capacitance and energy stored?

The capacitance of a parallel plate capacitor is given by C = ЄA/d, where Є is the permittivity of a medium kept inside the plates. 'A' be area of plates and 'd' is the distance between the plates of the capacitor. If two plates of a charged capacitor are brought closer to each other, 'd' will be decreased. As a result, capacitance 'C' increases.

2. Two identical capacitors are connected in series. Is the resulting capacitance greater or less than that of each individual capacitor?

Let two identical capacitors of capacitance C be connected in series, then the total capacitance is given by 1/C` = 1/C + 1/C, the net capacitance will be less than the individual capacitance.

3. What is the physical significance of relative the permittivity of a material placed between two plates of a capacitor?

The capacitance of parallel plate capacitor is given by;

C0 = Є0 A/d

Where A is the area of cross-section, d is the distance between plates and Є0 is the permittivity of the free space. If a dielectric is inserted between the plate, then capacitance is

Cm = Є A/ d

Then,

Cm/ C0 = Є/ Є0 = Єr,

where Єr is the relative permittivity, which is greater than 1. So, the capacitance of a capacitor increases when a dielectric is inserted between them.

Again q = C. V, this shows that potential decreases on increasing capacitance.

4. What are the factors that determine the value of capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is given by C = Є A/ d, where Є is the permittivity of the medium that is kept in between the plates if the capacitor, A is the common area of the plates and d is the separation between the plates. From this relation, we can see that the capacitance of a parallel plate capacitor is determined by the factors like area of plate A, distance between the plate d, and permittivity of medium Є.

5. Two parallel plate capacitors identical in shape and size are filled with air and mica separately. Which one has a larger capacitance? Write a proper relation in support of your answer.

The capacitor of parallel plate capacitor is given by,

C0 = Є0 A/ d

Where A is the area of cross-section, d is the distance between them and Є0 is permittivity of free space. If a dielectric is inserted between the plate, then capacitance is

Cm = Є/ Є0 = Єr

Where Єr is the relative permittivity, which is greater than 1. So, the capacitance of a capacitor increases when a dielectric is inserted between them.

The capacitor mica will have a larger capacity than the air capacitor because mica has a higher dielectric constant.

6. Explain the difference between dielectric constant and dielectric strength.

 Dielectric constant Dielectric strength Dielectric constant (K) of a medium is defined as the ratio of permittivity (Є) of the medium to permittivity (Є0) of free space. Dielectric strength of a medium is the maximum electric field intensity applied to it that can withstand by the medium without insulation breakdown. It has no unit Its SI unit is NC-1 or Vm-1 It is a pure number, so it is scalar. It is a vector.

7. Can we give any desired change to a capacitor? Explain.

No, we can’t give any desired change to a capacitor. We can charge the capacitor to its dielectric strength but not over its value. As we increase the charge to a capacitor, the potential difference between the plates also increases. A stage is reached when the electric field between the two plates attains the breakdown value of dielectric, the charges start to leak.

8. What are the uses of dielectrics in capacitors?

The capacitance of the parallel plate is given by

C0 = Є0A/ d

Where A is the area of cross-section, d is distance between them and Є0 is the permittivity of free space. If a dielectric is inserted between the plates, then capacitance is

Cm­= ЄA/ d

Then,

Cm­/C0 = Є/ Є0 = Єr

Where Єr is the relative permittivity, which is greater than 1. So, the capacitance of a capacitor increases when a dielectric is inserted between them.

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