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a 10.00mf parallel-plate capacitor is connected to a 24.0v battery. after the capacitor is fully charged ,the
battery is disconnected without loss of any of charge on the plates.
a) a voltmeter is connected across the two plates without discharging them.what does it read
b)what would the voltmeter read if the plate separationwere doubled?

Sagot :

The answer to the first part of the question is 24 V and the answer to the second part of the question is 48 V.

The formula which relates Charge, Capacitance and Voltage is

Charge = Capacitance × Voltage

Q = CV

where Q denotes the charge, C denotes the capacitance and V denote the voltage.

Using the above formula the charge on the capacitance will be

Q = CV

C = 10 mF

V = 24 V

Q = 240 mC

Charge on capacitance is 240 mC.

Now after we disconnect the battery

C = 10 mF (will remain same)

Q = 240 mC

V = Q/C

V = 240/10 V

V = 24 V

So after we removed the battery, The voltage will remain same.

Now we know that parallel plate capacitor formula is

C = ε(A/d)

from here

C ∝ 1/d, and as there is no loss of charge we can say that V ∝ 1/d

Form C ∝ 1/d and V ∝ 1/C we can state that

V ∝ d   {where d is the measurement of separation between the plates}

so if we double the distance between the plates then, the voltage will also get double.

Previously our voltage was 24 V, now if we double the distance between the plates the voltage will also get double, and it will become 48 V.

So, the voltmeter will take a reading of 24 V if voltmeter is connected across the two plates without discharging them, and if we double the plate separation then it will take 48 V as reading.

Learn more about Capacitors here:

https://brainly.com/question/13578522

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