Question

A certain parallel-plate capacitor is filled with a dielectric for which κ = 6.56. The area of each plate is 0.0830 m2, and the plates are separated by 1.95 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 202 kN/C. What is the maximum energy that can be stored in the capacitor?

Answer 1

Answer:

$4.86\times 10^{-7}\ J$

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 $(E_{max})$ = 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:

$V_{max}=E_{max}d\\\\V=(202000\ N/C)(0.00195\ m)\\\\V=393.9\ V$

Now, capacitance of the capacitor is given as:

$C=\dfrac{k\epsilon_0A}{d}\\\\C=\frac{6.56\times 8.854\times 10^{-12}\ F/m\times 0.0830\ m^2}{0.00195\ m}\\\\C=2.47\times 10^{-9}\ F$

The maximum energy stored in the capacitor is given as:

$U_{max}=\frac{1}{2}CV_{max}^2\\\\U_{max}=\frac{1}{2}\times (2.47\times 10^{-9}\ F)\times 393.9\ V\\\\U_{max}=4.86\times 10^{-7}\ J$

Therefore, the maximum energy that can be stored in the capacitor is $4.86\times 10^{-7}\ J$