Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electrostatic Potential and Capacitance

Question:

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 if dielectric slab with dielectric constant K=3 and same thickness were inserted in place of metal plate?

Options:

1535 V

1500 V

1452 V

1466 V

Correct Answer:

1466 V

Explanation:

C' = \(\frac{\epsilon_{o} A}{d - t(1-1/k)}\) 

k = 3 ; d = 0.05 mm ; t = 0.02 mm 

C' =  \(\frac{\epsilon_{o} A}{0.05)}\) \(\frac{15}{11}\)

Q is constant ∴ C' V' = C V 

V = E x d 

E = 4X104 V/m ; d = 0.05mm

V = 2000 V

V' = 1467.67 V