Electrostatics Ques 133
- Five identical capacitor plates, each of area $A$, are arranged such that adjacent plates are at a distance $d$ apart, the plates are connected to a source of emf $V$ as shown in the figure.
(1984, 2M)
The charge on plate 1 is and on plate 4 is ……
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Answer:
Correct Answer: 133.$\frac{\varepsilon_0 A V}{d},-\frac{2 \varepsilon_0 A V}{d}$
Solution:
Formula:
- In the circuit shown in figure, there is a capacitor between plates 1 and 2 , the capacity of which is : $C_{1}=\frac{\varepsilon_{0} A}{d}$ and potential difference between its plates is $V$. Therefore, charge stored in it is, $q=C_{1} V=\frac{\varepsilon_{0} A V}{d}$
Since, plate 1 is connected with positive terminal, hence this charge $q$ will be positive.
Plate 4 is making two capacitors, one with 3 and other with 5.
Hence, charge on it will be $-2 q$ or $\frac{-2 \varepsilon_{0} A V}{d}$. Charge on it is negative because this is connected with negative plate. Charges on both sides of the plates are shown below.
Here,
$$ q=\frac{\varepsilon_{0} A V}{d} $$
BETA
