Start Free Mock LoginAlready registered? Login here

Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorBasic capacitor analysis with dielectricMedium2 minQB

PhysicsMediumnumerical

A parallel plate capacitor has charge $5 \times 1 0^{- 6} C$ . A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is $4 \times 1 0^{- 6} C$ then the dielectric constant of the slab is ________.

Answer:

5.00

Solution:

$Q_{in} = Q (1 - \frac{1}{K}) 4 \times 1 0^{- 6} = 5 \times 1 0^{- 6} (1 - \frac{1}{K}) 1 - \frac{1}{K} = \frac{4}{5} K = 5$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Basic capacitor analysis with dielectric

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...