A block A of mass m kept on a rough horizontal surface is connected to a dielectric slab of mass m/6 and dielectric constant k by means of a light and inextensible string passing over a fixed pulley as shown in Fig. The dielectric can completely fill the space between the parallel plate capacitor of plate area λ × λ and the separation between plates d kept in vertical position. Initially switch S is open and length of the dielectric inside the capacitor is b. The coefficient of friction between the block A and the surface is µ = 1/4. Ignore any other friction.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. The forces acting on the dielectric are electrostatic attractive force of field of capacitor and its weight. The block will slip when
F E + mg 
µMg
F E 
g – 
g
(k – 1)V 2 
∴ V min = 
= 
.
Now V = 2V min. In this case the block will accelerate
Dielectric: F E + mg – T = ma….. (1)
Block: T – µMg = Ma ….. (2)
Eq. (1) and (2) give
a = 
As F E = 
(k – 1)V 2 = 
(k – 1) × 4 × 
= 2Mg
Thus a = 
= 
= 
g
From equation of motion, v
2 = 2as, we have
v 2 = 2 
× (l × b)
v = 
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