Question

The capacitance of a capacitor is 1000 µf. if the distance between the plates of the capacitor is halved and

the area of the plates are doubled, then the capacitance of the new capacitor is ____

a) 2000 µf b) 4000 µf c) 250 µf d) 500 µf​

Answer 1

Answer:

b is the correct answer 4000uf

Explanation:

C=kAE•/d = 1000uf

where k= constant of dielectric

A= area of of 2 plates

d=distance between 2 plates

E• = epsilon not = 8.85×10‐¹²

Now if we double the area and half the distance then the capacitance is given by

C'= k × 2A×E•×2 = 4 k E•A = 4 ×C

_______________ __________

d d

C'= 4×1000 = 4000uf