The plates of the parallel plate capacitor have plate area A and are clamped in the laboratory as shown in figure. The dielectric slab of mass m, length 2 λ and width 2 λ is released from rest with length λ inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period. (plates of capacitor are square plates of side 2 λ )
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(4. 
)
Sol. When dielectric slab is released from rest constant force act on slab towards the mean position after mean position same opposite force is act on slab which retard it come in rest position. Therefore motion of slab is periodic.
F = 
(K–1) V 2 , a = 
(K–1) V 2
by S = 
(acc.) t 2 , ( λ ) = 
(K–1) V 2 . t 2
t = 
⇒ t = 2 
⇒ T = 4t = 4 
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems