1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Consider a parallel-plate capacitor of capacitance 10 μF with air filled in the gap between the plates. Now, one half of the space between the plates is filled with a dielectric of dielectric constant 4 as shown in the figure. The capacitance of the capacitor changes to

A
25 μF
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
20 μF
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
20 μF
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
40 μF
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 25 μFFrom given we can redraw circuit as Ceq=C′1+C′2=Kε0A2d+ε0A2d=Kε0A2d+ε0A2d=ε0A2d[K+1] From given ε0A2d=10 μF and K=4. On substituting these values we get, Ceq=25 μF

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Placement of Dielectrics in Parallel Plate Capacitor
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program