Question

P.3-17 Solve the following problems: a) Find the breakdown voltage of a parallel-plate capacitor, assuming that e conducting plates are 50 (mm) apart and the medium between them is air. b) Find the breakdown voltage if the entire space between the conducting plates is filled with plexiglass, which has a dielectric constant 3 and a dielectric strength 20 (kV/mm). c) If a 10-(mm) thick plexiglass is inserted between the plates, what is the maximum voltage that can be applied to the plates. without a breakdown?

I dont understand how they solved the c part could someone help out..

Especially how do they got that formula i highlighted

Could someone explain this problem clearly and correctly..

Any helpful help would be appreciated..

Answer 1

Given that a find a break down voltage at a parallel plate capacitor assuming that conducting plates are 50 (mm) apart and the medium between then is air rightarrow from a table 3-1 in the text book consider dielectric strength of air is 3 times 10^6 V/m rightarrow consider the distance between the conducting plates 50 mm rightarrow consider the medium between the parallel plate capacitor is air rightarrow calculate the breakdown voltage of parallel plate capacitor v_b v_b = f_b a d_a = (3 times 10^6) (50 times 10^- 3) = 150 kv therefore the break down voltage v_b of parallel plate capacitor is 150 kv find the break down voltage if the entire space between conducting plates is filled with flexi glass, which has dielectric constant 3 and a dielectric strength 20 (kv/mm) consider the medium between the parallel plate capacitor is flexi glass consider dielectric constant 3 consider the break down voltage