If the PLATE SEPARATION of an isolated charged parallel-plate capacitor is doubled: A. the electric field is doubled

Physics

1 answer:

mash [69]3 years ago

5 0

Explanation:

The electric field of an isolated charged parallel-plate capacitor is given by :

$E=\dfrac{q}{A\epsilon_o}$ ........(1)

Where

q is the electric charge

A is the area of cross section of parallel plate

It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.

So, the correct option is (E) i.e. "none of the above".

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b) $emf = -d\Phi/dt = -\frac{\mu_0}{2\pi }.b.ln(\frac{a+d}{d}).(2.98t)= \\=-\frac{4\pi * 10^{-7}*0.027}{2\pi} ln(1.46/0.46)*2.98*2.90= -5.39 * 10^{-8} V$