Answer:
[tex]4.86\times 10^{-7}\ J[/tex]
Explanation:
Given:
Dielectric of the medium between the plates (k) = 6.56
Area of eac plate (A) = 0.0830 m²
Separation between the plates (d) = 1.95 mm = 0.00195 m [1 mm = 0.001 m]
Maximum electric field [tex](E_{max})[/tex] = 202 kN/C = 202000 N/C [1 kN = 1000 N]
Permittivity of space (ε₀) = 8.854 × 10⁻¹² F/m
The maximum potential difference across the plates of the capacitor is given as:
[tex]V_{max}=E_{max}d\\\\V=(202000\ N/C)(0.00195\ m)\\\\V=393.9\ V[/tex]
Now, capacitance of the capacitor is given as:
[tex]C=\dfrac{k\epsilon_0A}{d}\\\\C=\frac{6.56\times 8.854\times 10^{-12}\ F/m\times 0.0830\ m^2}{0.00195\ m}\\\\C=2.47\times 10^{-9}\ F[/tex]
The maximum energy stored in the capacitor is given as:
[tex]U_{max}=\frac{1}{2}CV_{max}^2\\\\U_{max}=\frac{1}{2}\times (2.47\times 10^{-9}\ F)\times 393.9\ V\\\\U_{max}=4.86\times 10^{-7}\ J[/tex]
Therefore, the maximum energy that can be stored in the capacitor is [tex]4.86\times 10^{-7}\ J[/tex]
