Questions
A potential difference of 2000 V is established between the parallel plates of a capacitor with air between the plates. If the air becomes conducting when the electric field exceeds 3 x 106 N/C, the minimum separation between the plates should be
detailed solution
Correct option is C
Let the minimum separation should be x. Then E=Vd=2000x=3×106∴ x=20003×106=23×10−3=0.667mmTalk to our academic expert!
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Two identical metal plates are given positive charges Q1 and Q2 (<Q1) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is
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