Consider a parallel plate capcaitor with plates 20 cm by 20 cm and separated by 2 mm. The dielectric constant of the material between the plates is 5. The plates are connected to a voltage source of 500 V. The energy density of the field between the plates will be close to

Options

A. `2.65J//m^(3)`
B. `1.95J//m^(3)`
C. `1.38J//m^(3)`
D. `0.69J//m^(3)`

Answer

C

Explanation

Electric field density density or energy per unit volume is ` u =(k epsi_(0))/(2) ((V)/(d))^(2)`
` k = 5, V = 500` Volts `d = 2 xx 10^(-3)m`
`therefore u = (5)/(2) xx(1)/(4pi xx 9 xx 10^(9)) xx ((500)^(2))/((2xx10^(-3))^(2)) =1.38J m^(-3)`