Sarthaks Test
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(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. 

(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor.

(c) Two identical point charges, q each, are kept 2 m apart in air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q.

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(a).The plates have been changed with Q and -Q , 

By connecting them to the two terminals of a battery. 

Q is called the charge of the capacitor, 

Through the total charge of the capacitor is zero.

Intervening medium : air/vaccum

Let us consider insulated metal plate A, 

Charged to Q, till its potential becomes maximum. Bring another insulated metal plate B near to plate A. By induction, a negative charge is produced on inner surface of B and equal positive charge on farther face as shown in figure. 

The induced negative charge tends to decrease the potential of A and induced positive charge tends to increase the potential of A. 

The overall potential of A reduces and hence some more charge can be given to A to raise its potential to the same maximum value. 

Thus, 

Capacity of conductor A has increased, by bringing another uncharged conductor B in its vicinity. Earthing B, makes positive charge flew to earth.

Thus, 

Due to the induced negative charge on B, potential of A is greatly reduced. 

Thus, 

A large amount of charge can be given to A to raise it to the maximum potential.

(b) Charge on the capacitor(single), 

Q = CV, 

Where C is the capacitance of both the capacitors(common).

∴ Energy stored in the single capacitor,

E1 = \(\frac{1}{2}cv^2\) 

\(\frac{1}{2}Qv\) 

Now, 

In 2nd case, 

When another uncharged capacitor of capacitance C, is connected to the single capacitor, then they reach common potential, V'.

∴ Now, 

The charge on each capacitor, 

Q' = CV'

By charge conservation,

Q' = \(\frac{Q}{2}\) 

⇒ V' = \(\frac{V}{2}\) 

∴ Energy sotred in the combination, 

E1 = \(\frac{1}{2}\times\) 2 \(\times\) Q' \(\times\) V'

\(\frac{Q}{2}\)\(\times\)\(\frac{V}{2}\) 

\(\frac{1}{4}\)QV

(c) 

For the charge +q at A to be in equilibrium, 

Force on A due to Q at B+ Force on A due to Q at O = 0

\(\frac{1}{4\pi ∈_0}.\frac{q^2}{Z^2}\)

\(\frac{1}{4\pi ∈_0}.\frac{q\times Q}{1^2}\) 

Q = \(\frac{q^2}{4q}\) 

\(\frac{q}{4}\)

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