First slide

Capacitance

Question

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

Easy

A

2E
B

E
C

E/4
D

E/2

Solution

Here, the charge on the capacitor is constant, i.e.,
q = constant
$∴ CV = constant$
or $\frac{ε_{0} A}{d} \times V = constantor \frac{V}{d} = constant$
$∴ E = constant$

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