Question

NAVNEET 21 MOST LIKELY QUESTION SETS : PHYSICS-STD. XII
24. A cube of amber, 1 cm on the side, is kept in an electric field of
intensity 200 V/m. Determine the electrostatic energy contained in
the cube of amber. [Dielectric constant of amber = 2.8] (3)​

Answer 1

Answer:

Electrostatic field energy stored in the cube is

$U = 4.96 \times 10^{-13} J$

Explanation:

Energy stored due to electric field region in given as

$U = \frac{1}{2}\epsilon_0 E^2 (Volume)$

here we know that the medium is dielectric so we will have

$U = \frac{1}{2}k\epsilon_0 E^2 V$

now we know that

k = 2.8

E = 200 V/m

length of side of the cube is 1 cm

so we will have

$U = \frac{1}{2}(2.8)(8.85 \times 10^{-12})(200^2)(0.01)^3$

$U = 4.96 \times 10^{-13} J$

#Learn

Topic : electrostatic field Energy

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