Question

A capacitor has two parallel plates seperated by 2mm and is connected across a 50V battery. a. What is the electric field between the field? b. What is the surface charge density? c. How much charge is stored on each plate if the area is 0.1m^2. d. Calculate the capicitance e. How much energy is stored in this capacitor?

Answer 1

a

`E =  25000 \  V/m`

b

`\sigma  =  2.2125* 10^(-7) \  C/m^2`

c

`q = 2.2125 *10^(-8) \  C`

d

`C = 4.425 *10^(-10) \ F`

e

`U  = 5.531 *10^(-7) \  J`

Step-by-step explanation:

From the question we are told that

The distance of separation is
`d =  2mm =  0.002 \ m`

The voltage is
`V  = 50\  V`

The area is
`A  =0.1m^2`

Generally the electric field is mathematically represented as

`E =  (V)/(d)`

=>
`E =  (50)/( 0.002)`

=>
`E =  25000 \  V/m`

Generally the capacitance mathematically represented is

`C =  (\epsilon_o  * A  )/(d)`

Here
`\episilon_o` is the permitivity of free space with value

`\episilon_o = 8.85 * 10^(-12) C/(V⋅m`

=>
`C =  ( 8.85 * 10^(-12)  * 0.1  )/(0.002)`

=>
`C = 4.425 *10^(-10) \ F`

Generally the charge is mathematically represented as

`q =  C  * V`

`q = 4.425 *10^(-10)   *50`

=>
`q = 2.2125 *10^(-8) \  C`

Generally the charge density is mathematically represented as

`\sigma  =  (q)/(A)`

=>
`\sigma  =  ( 2.2125 *10^(-8))/(0.1)`

=>
`\sigma  =  2.2125* 10^(-7) \  C/m^2`

Generally the energy stored in this capacitor is mathematically

`U  = (1)/(2) *  CV^2`

=>
`U  = (1)/(2) *  (4.425 *10^(-10) ) * (50)^2`

=>
`U  = 5.531 *10^(-7) \  J`