Respuesta :
Answer:
The stored energy is [tex]3.83\times10^{-6}\ J[/tex].
Explanation:
Given that,
One side of square plates = 8.5 cm
Distance = 2.5 mm
Charge = 14 nC
We need to calculate the energy stored
The energy is stored by the electric field between two square plates is defined as,
[tex]U=\dfrac{1}{2}\times\epsilon_{0}\times(\dfrac{q}{A\epsilon_{0}})^2\times A\times D[/tex]
[tex]U=\dfrac{1}{2}\times\dfrac{q^2}{A\epsilon_{0}}\times D[/tex]
We calculate the area iof square plates
[tex]A=(8.5\times10^{-2})^2[/tex]
[tex]A=0.007225=72.2\times10^{-4}\ m^2[/tex]
Now, put the value into the formula
[tex]U=\dfrac{1}{2}\times\dfrac{(14\times10^{-9})^2}{72.2\times10^{-4}\times8.85\times10^{-12}}\times2.5\times10^{-3}[/tex]
[tex]U=3.83\times10^{-6}\ J[/tex]
Hence, The stored energy is [tex]3.83\times10^{-6}\ J[/tex].
The quantity of energy that is stored by the electric field between two square plates is [tex]3.83\times 10^{-6}\;Joules[/tex]
Given the following data:
- Distance = 2.5 mm to m = 0.0025 meter
- Distance on plate = 8.5 cm to m = 0.085 meter
- Charge = 14 nC = [tex]14\times10^9\;C[/tex]
Scientific data:
Vacuum permittivity = [tex]8.85 \times 10^{-12}\;F/m[/tex]
To determine the quantity of energy that is stored by the electric field between two square plates:
First of all, we would determine the area of the square plates.
[tex]Area = s^2\\\\Area = (0.085)^2\\\\Area = 7.225 \times 10^{-3} \;m^2[/tex]
Mathematically, the quantity of energy that is stored in an electric field between two square plates:
[tex]U=\frac{1}{2} \frac{q^2D}{A \epsilon_o}[/tex]
Where:
- U is the energy stored in an electric field.
- D is the distance.
- q is the charge.
- [tex]\epsilon_o[/tex] is the vacuum permittivity.
Substituting the given parameters into the formula, we have;
[tex]U=\frac{1}{2} \times \frac{(14\times 10^{-9})^2 \; \times \;0.0025}{7.225 \times 10^{-3} \;\times \;8.85 \times 10^{-12} }\\\\U=\frac{1}{2} \times \frac{(1.96\times 10^{-16}) \; \times \;0.0025}{7.225 \times 10^{-3} \;\times \;8.85 \times 10^{-12} }\\\\U=\frac{1}{2} \times \frac{4.9\times 10^{-19} }{6.39 \times 10^{-14} }\\\\U = 3.83\times 10^{-6}\;Joules[/tex]
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