Question

Two isolated copper plates, each of area 0.40 m2, carry opposite charges of magnitude 7.08 × 10-10 C. They are placed opposite each other in parallel alignment, with a spacing of 4.0 cm between them. What is the potential difference between the plates? (ε0 = 8.85 × 10-12 C2/N ∙ m2)

Answer 1

The potential difference between the plates is 8 V.

Step-by-step explanation:

Given that,

Area of plates = 0.40 m²

Charge
`q=7.08*10^(-10)\ C`

Distance = 4.0 cm

We need to calculate the electric field

Using for formula of electric field

`E=(2q)/(2\epsilon_(0)A)`

Where, q = charge

A = area

Put the value into the formula

`E=(7.08*10^(-10))/(8.85*10^(-12)*0.40)`

`E=200\ V/m`

We need to calculate the potential difference between the plates

Using formula of potential difference

`V=E* d`

Where, E = electric field

d = distance

Put the value into the formula

`V=200*0.04`

`V=8\ V`

Hence, The potential difference between the plates is 8 V.