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julsineya [31]
3 years ago
10

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

electric field between the two plates?
Physics
1 answer:
Stells [14]3 years ago
8 0

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

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Find the total electric charge of 2.5 kg of electrons. Express your answer using two significant figures.
zimovet [89]

Answer : The total electric charge of electrons is, -4.4\times 10^{11}C

Explanation:

Answer : The number of electrons transferred are, 4.68\times 10^{20}

Explanation :

First we have to calculate the number of electrons.

Number of electrons = \frac{\text{Total mass of electrons}}{\text{Mass of one electron}}

Mass of 1 electron = 9.1\times 10^{-31}kg

Total mass of electron = 2.5 kg

Number of electrons = \frac{2.5kg}{9.1\times 10^{-31}kg}

Number of electrons = 2.75\times 10^{30}

Now we have to calculate the total electric charge of electrons.

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Q=ne\\\\n=\frac{Q}{e}

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