Question

Each of the plates shown in figure has surface
area A on one side and the separation between the
consecutive plates is d. The emf of the battery
connected is E. The magnitude of charge
transferred through the battery is​

Answer 1

The magnitude of charge  transferred through the battery is $\bold{\frac{\epsilon_0 AE}{3}}$

Given:

Surface area = A

Distance between the plates = d

Battery emf = E

To find:

Magnitude of charge = Q = ?

Solution:

Capacitance of the capacitor is given by the formula:

$\bold{C=\frac{\epsilon_0 A}{d}}$

Since, the capacitors are connected in series.

$\bold{C_{series}=\frac{C}{3}=\frac{\epsilon_0A}{3d}}$

Now, the relationship between capacitance and charge is given by the formula:

$\bold{Q=CV}$

$\bold{Q=\frac{\epsilon_0A}{3d}\times Ed}$

$\bold{\therefore Q= \frac{\epsilon_0 AE}{3}}$

Answer 2

Answer:

answer is 4th option

Explanation:

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