Question 25.47: A certain substance has a dielectric constant of 2.8 and a d......
A certain substance has a dielectric constant of 2.8 and a dielectric strength of 18 MV/m. If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of 7.0 × 10^{-2} μF and to ensure that the capacitor will be able to withstand a potential difference of 4.0 kV?
THINK Dielectric strength is the maximum value of the electric field a dielectric material can tolerate without breakdown.
EXPRESS The capacitance is given by C=\kappa C_0=\kappa \varepsilon_0 A / d , where C_0 is the capacitance without the dielectric, \kappa is the dielectric constant, A is the plate area, and d is the plate separation. The electric field between the plates is given by E = V/d, where V is the potential difference between the plates. Thus, d = V/E and C=\kappa \varepsilon_0 A E / V . Therefore, we find the plate area to be
A=\frac{C V}{\kappa \varepsilon_0 E} .
ANALYZE For the area to be a minimum, the electric field must be the greatest it can be without breakdown occurring. That is,
A=\frac{\left(7.0 \times 10^{-8} \,F \right)\left(4.0 \times 10^3\, V \right)}{2.8\left(8.85 \times 10^{-12} \,F / m \right)\left(18 \times 10^6\, V / m \right)}=0.63 \,m ^2 .
LEARN If the area is smaller than the minimum value found above, then electric breakdown occurs and the dielectric is no longer insulating and will start to conduct.