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.

Question

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.

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

Monjur Alahi

11 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
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)`

11 views Answered 2023-02-05 15:29