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

Answer:

The potential difference between the plates is 8 V.

Explanation:

Given that,

Area of plates = 0.40 m²

Charge $q=7.08\times10^{-10}\ C$

Distance = 4.0 cm

We need to calculate the electric field

Using for formula of electric field

$E=\dfrac{2q}{2\epsilon_{0}A}$

Where, q = charge

A = area

Put the value into the formula

$E=\dfrac{7.08\times10^{-10}}{8.85\times10^{-12}\times0.40}$

$E=200\ V/m$

We need to calculate the potential difference between the plates

Using formula of potential difference

$V=E\times d$

Where, E = electric field

d = distance

Put the value into the formula

$V=200\times0.04$

$V=8\ V$

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