Physics, asked by sejalbharambe2003, 10 months ago

Q. 26. A cube of amber, 1 cm on the side, is kept in an electrostatic field of inter
200 V/m. Determine the energy contained in the cube of amber.
[Dielectric constant of amber = 2.8]​

Answers

Answered by mihirvekariya2002
1

Answer:

1.982×10^-12 Joule

b.

Answered by CarliReifsteck
0

The energy contained in the cube of amber is 4.956\times10^{-13}\ J

Explanation:

Given that,

Side of cube = 1 cm

Dielectric constant of amber= 2.8

Electrostatic field = 200 V/m

We need to calculate the energy contained in the cube of amber

Using formula of energy

U=\dfrac{1}{2}K\epsilon_{0}E^2V

Where, V = volume

K = Dielectric constant

E = Electrostatic field

Put the value into the formula

U=\dfrac{1}{2}\times2.8\times8.85\times10^{-12}\times(200)^2\times10^{-6}

U=4.956\times10^{-13}\ J

Hence, The energy contained in the cube of amber is 4.956\times10^{-13}\ J

Learn more :

Topic : electrostatic field

https://brainly.in/question/1740669

Similar questions