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azamat
3 years ago
9

An electric field of 8.5X10^5 V/m is desired between two parallel plates each of areas 2500 cm^2 and separated by 0.10 mm of air

. What charge must be on each plate?
Physics
1 answer:
aksik [14]3 years ago
6 0

Answer:

The charge must be on each plate is 1.881\times10^{-6}\ C.

Explanation:

Given that,

Electric field E= 8.5\times10^{5}\ V/m

Area = 2500 cm²

Distance = 0.10 mm

We need to calculate the potential difference

Using formula of potential difference

\Delta V=E\times d

\Delta V=8.5\times10^{5}\times0.10\times10^{-3}

\Delta V=85\ V

We need to calculate the capacitor

Using formula of capacitor

C=\dfrac{\epsilon_{0}\times A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times2500\times10^{-4}}{0.10\times10^{-3}}

C=2.2125\times10^{-8}\ F

We need to calculate the charge

Using formula of charge

Q=C\times\Delta V

Q=2.2125\times10^{-8}\times85

Q=1.881\times10^{-6}\ C

Hence, The charge must be on each plate is 1.881\times10^{-6}\ C.

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