Question

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]​

Answer 1

Answer:

1.982×10^-12 Joule

b.

Answer 2

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 $4.956\times10^{-13}\ J$

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Topic : electrostatic field

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