Complete Question
Two parallel circular plates with radius 7 mm carrying equal -magnitude surface charge densities of ± 3.0μC/m2 are separated by a distance of 1 mm. How much stored energy do the plates have? (ε0 = 8.85 × 10-12 C2/N ∙ m2)
a. 7.9 nJ b. 25 nJ c. 78 nJ d. 250 nJ
Answer:
The energy stored in the plate is 78 nJ
Explanation:
From the question we are given that
The radius is 7 mm [tex]= 0.7*10^{-3} \ m[/tex]
The surface charge density [tex]\sigma[/tex] [tex]= 3.0\mu C / m^2[/tex]
The distance between the plates as d = 1mm [tex]= 1 *10^{-3}m[/tex]
Generally the formula for energy stored in a capacitor is mathematically given as
[tex]U = \frac{\sigma^2* A*d}{2 \epsilon_0}[/tex]
Where A is the area which is mathematically
= [tex]\pi r^2[/tex]
Substituting values into the equation we have
[tex]U = \frac{\sigma^2 (\pi r^2)d}{2 \epsilon_0}[/tex]
[tex]U = \frac{(3,0*10^{-6})^2 [(3.142)(7.0*10^{-3})^2] *(1.0*10^{-3})}{(2 *(8,85*10^{-12}))} =78nJ[/tex]
The energy stored in each plate of the capacitor is 78 nJ.
The energy stored in parallel plate capacitors is calculated as follows;
[tex]U = \frac{\sigma^2 Ad}{2\varepsilon _o} \\\\}[/tex]
where;
The energy stored in the plates is calculated as follows;
[tex]U = \frac{\sigma^2 (\pi r^2) d}{2 \varepsilon _o} \\\\U = \frac{(3\times 10^{-6})^2 \times \pi \times (7\times 10^{-3})^2 \times 1 \times 10^{-3} }{2 \times 8.85\times 10^{-12} } = 78 \times 10^{-9} J\\\\U = 78 \ nJ[/tex]
Thus, the energy stored in each plate of the capacitor is 78 nJ.
The complete question is below:
Two parallel circular plates with radius 7 mm carrying equal -magnitude surface charge densities of ± 3.0μC/m2 are separated by a distance of 1 mm. How much stored energy do the plates have? (ε0 = 8.85 × 10-12 C2/N ∙ m2)
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