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2 years ago

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Two plates of area 30.0 cm^2 are separated by a distance of 0.0590 cm. If a charge separation of 0.0240 μC is placed on the two

plates, 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:

topjm [15]2 years ago

3 0

Answer:

V=533.33 V

Explanation:

Given that

A= 30 cm²

d= 0.059 cm

Q= 0.0240 μ C

We know that capacitance given as

$C=\dfrac{\varepsilon _oA}{d}$

Now by putting the values

$C=\dfrac{8.85\times 10^{-12}\times 30\times 10^{-4}}{0.059\times 10^{-2}}$

$C=4.5\times 10^{-11}\ F$

Voltage difference given as

Q= V .C

V=Q/C

$V=\dfrac{0.0240\times 10^{-6}} {4.5\times 10^{-11}}$

V=533.33 V

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