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babymother [125]

3 years ago

7

Two plates of area 7.00 × 10-3 m2 are separated by a distance of 2.60 × 10-4 m. If a charge of 5.40 × 10-8 C is moved from one p

late 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.

1 answer:

lesantik [10]3 years ago

6 0

Answer:

226.53 Volt

Explanation:

A = Area of plates = 7.00×10⁻³ m²

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

d = Distance between two plates = 2.60×10⁻⁴ m

Q = Charge = 5.40×10⁻⁸ C

Capacitance

$C=\frac{\epsilon_0A}{d}\\\Rightarrow C=\frac{8.854\times 10^{-12}\times 7\times 10^{-3}}{2.6\times 10^{-4}}\\\Rightarrow C=23.83\times 10^{-11}$

Potential difference between plates

$V=\frac{Q}{C}\\\Rightarrow V=\frac{5.4\times 10^{-8}}{23.83\times 10^{-11}}\\\Rightarrow V=226.53\ Volt$

∴ The potential difference (voltage) between the two plates is 226.53 Volt

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