Respuesta :
Explanation:
The given data is as follows.
Dielectric constant, K = 3.0
Area of the plates (A) = 0.021 [tex]m^{2}[/tex]
Distance between plates (d) = [tex]2.75 \times 10^{-3} m[/tex]
Maximum electric field (E) = [tex]3.25 \times 10^{5} V/m[/tex]
Now, we will calculate the capacitance as follows.
C = [tex]\frac{k \epsilon_{o} \times A}{d}[/tex]
= [tex]\frac{3.0 \times 8.85 \times 10^{-12} \times 0.021}{2.75 \times 10^{-3}}[/tex]
= [tex]\frac{0.55755 \times 10^{-12}}{2.75 \times 10^{-3}}[/tex]
= [tex]0.203 \times 10^{-9}[/tex] F
Formula to calculate electric charge is as follows.
E = [tex]\frac{\sigma}{k \epsilon_{o}}[/tex]
or, Q = [tex]E \times k \times \epsilon_{o}A[/tex] (as [tex]\frac{\sigma}{\epsilon_{o}} = \frac{Q}{A}[/tex])
= [tex]3.25 \times 10^{5} \times 3.0 \times 8.85 \times 10^{-12} \times 0.021[/tex]
= [tex]181.2 \times 10^{-9} C[/tex]
Formula to calculate the energy is as follows.
U = [tex]\frac{1 \times Q^{2}}{2 \times C}[/tex]
= [tex]\frac{(181.2 \times 10^{-9} C)^{2}}{2 \times 1.6691 \times 10^{-9}}[/tex]
= [tex]\frac{32833.44 \times 10^{-18}}{3.3382 \times 10^{-9}}[/tex]
= [tex]9835.67 \times 10^{-9}[/tex]
or, = [tex]98.35 \times 10^{-7} J[/tex]
Thus, we can conclude that the maximum energy that can be stored in the capacitor is [tex]98.35 \times 10^{-7} J[/tex].
