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Ksivusya [100]
4 years ago
12

Two isolated copper plates, each of area 0.40 m2, carry opposite charges of magnitude 7.08 × 10-10 C. They are placed opposite e

ach 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)
Physics
1 answer:
polet [3.4K]4 years ago
3 0

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.

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