Question

the capacitance of a variable capacitor joined with a battery of 100v is changed from 2 micro Faraday to 10 micro Faraday. what is the change in the energy stored in it ​

Answer 1

The capacitance of a variable capacitor joined with a battery of 100V is changed from 2 mF to 10 mF.

We have to find the change in the energy stored in capacitor.

As per given condition, there are two capacitors. One of capacitance 2mF and another of capacitance 10mF.

C1 = 2 mF and C2 = 10mF

Firstly, convert the mF into F. To do this, multiply the given value by 10^-6.

So,

C1 = 2 × 10^-6 F and C2 = 10 × 10^-6 F

Energy stored in capacitor is given by:

∆E = 1/2 CV²

As there are two capacitors. So,

∆E = 1/2 (C2 - C1) × V²

Substitute value of C1 = 2×10^-6, C2 = 10×10^-6 and V = 100V

∆E = 1/2 × (10 × 10^-6 - 2 × 10^-6) × (100)²

∆E = 1/2 × 10^-6 (10 - 2) × 10000

∆E = 1/2 × 10^(-6+4) × 8

∆E = 4 × 10^-2

Therefore, the change in energy stored in capacitor is 4 × 10^-2 J.

Additional Information

→ Ability to do work is known as Energy.

→ It's S.I. unit is joules (J).

→ Dimensional formula is [M¹L²T-²]

→ It is a scaler quantity. As it has magnitude only.

Answer 2

Aɴꜱᴡᴇʀ :-

Given

Battery of 100V changed from 2μF to 10μF

Nᴇᴇᴅ Tᴏ Fɪɴᴅ:-

The change in the energy stored in the capacitor

Sᴏʟᴜᴛɪᴏɴ:-

C₁ = 2μF

C₂ = 10μF

Firstly, converting μF → F to do this we have to multiply with 10-⁶

C₁ = 2 × 10⁻⁶

C₂ = 10 × 10⁻⁶

Finding C

C = C₂ - C₁

→ C = 10⁻⁵ - 2 × 10⁻⁶

→ C = 10⁻⁶ ( 10 - 2 )

→ C = 8 × 10⁻⁶

Energy stored in capacitor

∆E = ½ CV²

→ ∆E = 1/2 ( 8 × 10-⁶) ( 100 )²

→ ∆E = 1/2 × 8 × 10⁻⁶⁺⁴

→ ∆E = 4 × 10⁻²

.°. Change in energy stored in the capacitor = 4 × 10⁻²