Physics

Electrostatic Potential and Capacitance

Combination of capacitors

Five identical conducting plates 1,2,3,4 and 5 are fixed paral

Question

Hard

Solving time: 4 mins

Five identical conducting plates $1, 2, 3, 4$ and 5 are fixed parallel to and equidistant from each other as shown in fig. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 and the plate 4 are connected to a source of constant emf $V_{0}$ . Find : (i) The effective capacity of the system between the terminals of the source. (ii) The charge on plates 3 and 5 . Given $d =$ distance between any two successive plates and $A =$ area of either face of each plate.

Uploaded: 2023-10-02T11:44:02.000Z
Duration: 5 min

Views: 6,456 students

Updated on: Oct 2, 2023

Book:

Electrostatics (Disha)

Text solutionVerified

The equivalent arrangement of the plates is shown in Fig. From these five plates, there forms four identical capacitors. These are

C_{12}

C_{32}, C_{34}

and

C_{54}

. The capacitance of each capacitor

C = \frac{\in _{0} A}{d}

Capacitors

C_{12}

and

C_{32}

are in parallel and so their capacitance becomes

2 C

. Now

2 C

is in series with capacitor

C_{54}

, and so equivalent capacitance becomes

\frac{2 C}{3}

. And this now is in parallel with

C_{34}

. Thus effective capacitance between the terminals becomes

C_{e q} = C + \frac{2 C}{3} = \frac{5 C}{3}

Charge on the capacitor

C_{34}

Q_{34} = C V_{0}

. Charge on the capacitors

C_{54}, C_{12}

and

C_{32}

Q_{54} = \frac{2 C}{3} V_{0}

This charge equally divided on to

C_{12}

and

C_{32}

, so

Q_{12} = Q_{32} = Q = \frac{C V _{0}}{3}

The charge on plate

1 : Q_{1} = + Q_{12} = + \frac{C V _{0}}{3}

The charge on plate

2 : Q_{2} = - (Q_{12} + Q_{32})

= - 2 \frac{C V _{0}}{3} = - 2 \frac{\in _{0} A V _{0}}{3 d}

The charge on plate

3 : Q_{3} = + (Q_{32} + Q_{34}) = (C V_{0} + \frac{C V _{0}}{3})

= \frac{4}{3} C V_{0} = \frac{4}{3} \frac{\in _{0} A V _{0}}{d}

The charge on plate 4 :

Q_{4} = - (Q_{34} + Q_{54}) = - (C V_{0} + \frac{2 C V _{0}}{3}) = - \frac{5 C V _{0}}{3} = - \frac{5 \in _{0} A V _{0}}{3 d}

The charge on plate

5 : Q_{5} = + Q_{54}

= \frac{2 C V _{0}}{3} = \frac{2 \in _{0} A V _{0}}{3 d}

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Luna Discussed

Five identical conducting plates

1, 2, 3, 4

and 5 are fixed parallel to and equidistant from each other as shown in fig. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 and the plate 4 are connected to a source of constant emf

V_{0}

. Find : (i) The effective capacity of the system between the terminals of the source. (ii) The charge on plates 3 and 5 . Given

d =

distance between any two successive plates and

A =

area of either face of each plate.

7 mins ago

Discuss this question LIVE

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Question Text	Five identical conducting plates $1, 2, 3, 4$ and 5 are fixed parallel to and equidistant from each other as shown in fig. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 and the plate 4 are connected to a source of constant emf $V_{0}$ . Find : (i) The effective capacity of the system between the terminals of the source. (ii) The charge on plates 3 and 5 . Given $d =$ distance between any two successive plates and $A =$ area of either face of each plate.
Updated On	Oct 2, 2023
Topic	Electrostatic Potential and Capacitance
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	225
Avg. Video Duration	5 min

Text solutionVerified

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