Question

A cute of marble of each side 5 m long is placed in a uniform electric field Of intensity 3 V/m. the amount of electrostatic energy stored in marble of dielectric constant 8 is.

• A. $0.0399J$
• B. $1.366J$
• C. $0.75J$
• D. $2.3336J$

Answer 1

The correct option is A $0.0399J$

Electrostatic energy stored per unit volume is given by the following relation$,$

$\frac{W}{V}=\frac{1}{2}\in E^2$

$=\frac{1}{2}\left(\in {\in }_0\right)E^2$

$=\frac{1}{2}\left(8\right)\left(8.85\times {10}^{-12}\right){\left(3\times {10}^3\right)}^{2}J/m^3$

$=3.19\times {10}^{-4}J/m^3$

Now volume of cube is$,$

$V={\left(5m\right)}^3$

$=125m^3$

$W=\left(3.19\times {10}^{-4}J/m^3\right)\left(125m^3\right)$

$=0.0399J$

Hence,

option $\left(A\right)$ is correct answer.