Two square metal plates of side 1 m are kept 0.01 m apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of 500 V. The plates are then lowered vertically into the oil at a speed of 0.001 ms –1 . The current n×10 –9 A drawn from the battery during the process. Then find the value of n. (Dielectric constant of oil = 11], ( ∈ 0 = 8 × 10 –12 C 2 N –1 m –1 )
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(4)
Sol. Let a be the side of the square plate.
As shown in figure, C 1 and C 2 are in parallel. Therefore, total capacity of capacitors in the position shown is
C = C 1 + C 2
C = 
∴ q = CV = 
(a – x + Kx)
As plates are lowered in the oil, C increases hence charge stored will increase.
Therefore, i = 
= 
(K–1) · 
Substituting the values
= 8 × 10 –12 C 2 /N–m 2
a = 1m, V = 500 volt, d = 0.01m, K = 11 and 
= speed of plate = 0.001 m/s
We get current
i = 
Amp.
i = 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