Questions
A parallel plate condenser having a plate separation of d is charged to a potential V. It is then isolated. The intensity of electric field between the plates is then found to be E. The separation between the plates is doubled. The new electric field intensity is
detailed solution
Correct option is B
Here, the charge on the capacitor is constant, i.e., q = constant∴ CV= constant or ε0Ad×V= constantor Vd= constant ∴ E= constantTalk to our academic expert!
Similar Questions
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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