Case : Read the passage and answer the question(s).
In a parallel plate air capacitor having plate separation 0.05mm, an electric field of 4X104 V/m is established between the plates. After the removal of the battery a metal plate of thickness t = 0.02 mm is inserted between the plates of the capacitor.
What is the potential difference across capacitor after the introduction of metal plates?
Answer & explanation
Correct answer: option 2
V = E x d
E = 4 x 104 V/m ; d = 0.05 mm
V = 2000 V
Q = C V = 2000 C ;
C' = C' = \(\frac{C}{1 - t/d}\) = \(\frac{C}{1 - 0.02/0.05}\) = \(\frac{5}{3}\) C ;
Also, Q = C' V' = \(\frac{5}{3}\) C V' ;
C V = C' V' ⇒ V' = 1200 V