wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A capacitor, as shown in figure has square plates of length l and are inclined at an angle θ with one another. For small value of θ , capacitance is given by:


11481.png

A
ϵ0l2d(1θl2d)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
ϵ0l22d(1θld)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
ϵ0l2d(1+θld)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
ϵ0l22d(1+θld)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D ϵ0l2d(1θl2d)
At a distance x on the inclined plate, let us consider a small segment of length dx.
The small segment is at a distance of D from lower plate.
So, capacitance due to the small segment:
dC=ϵ0dAD=ϵ0dxl(d+xθ) as D=d+xsinθ and sinθθ for small θ
Integrating above, we get:
Cnet=l0ϵ0dAD=ϵ0lθl0θdx(d+xθ)=ϵ0lθ×ln(d+lθ)d
=ϵ0lθ×ln(1+lθd)
Using logarithm expansions,
Cnet=ϵ0lθ(lθd12×(lθ)2d2)
=ϵ0l2d(1θl2d)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Capacitance of a Parallel Plate Capacitor
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon