The electric field between the plates of a parallel–plate capacitance 2.0 µF drops to one third of its initial value in 4.4 µs when the plates are connected by a thin wire. Find the resistance of the wire in Ω. (Given: λ n3 = 1.0986)
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
( 
= 2.0 Ω.)
Sol. E = E 0 
⇒ 
= E 0 
= 
λ n 3 = 
⇒ R = 
= 2.0 Ω.
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
The capacity of a parallel plate condenser is . When a glass plate is placed between the plates of…
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then sli…
The energy of a charged capacitor is given by the expression ( = charge on the conductor and C = i…
The capacity of a condenser is farad and its potential is . The energy released on discharging it…
The insulated spheres of radii and having charges and respectively are connected to each other.…
In a charged capacitor, the energy resides