Question

Two plates with area 7.00×10−3 m27.00×10−3 m2 are separated by a distance of 4.80×10−4 m4.80×10−4 m . If a charge of 5.40×10−8 C5.40×10−8 C is moved from one plate to the other, calculate the potential difference (voltage) between the two plates. Assume that the separation distance is small in comparison to the diameter of the plates.

Answer 1

The voltage is
`V = 418.60 \ Volts`

Step-by-step explanation:

From the question we are told that

The area of the both plate is
`A = 7.00 *10^(-3) \ m^2`

The distance between the plate is
`d = 4.80*10^(-4)\ m`

The magnitude of the charge is
`q = 5.40 *10^(-8) \ C`

The capacitance of the capacitor that consist of the two plates is mathematically represented as

`C = (\epsilon _o A)/(d)`

Where
`\epsilon_o` is the permitivity of free space with a value
`e = 8.85*10^(-12) \ m^(-3) \cdot kg^(-1)\cdot s^4 \cdot A^2`

So

`C = (8.85*10^(-12) * (7* 10^(-3)))/( 4.8*10^(-4))`

`C = 1.29 *10^(-10) \ F`

The potential difference between the plate is mathematically represented as

`V = ( Q)/(C )`

`V = ( 5.4*10^(-8))/(1.29 *10^(-10))`

`V = 418.60 \ Volts`