IIT JEE , AIEEE Forum - Physics , Electricity

reddevil_2009

`The plates of capacitor,fully charged are conected through a resistor.Will its presence affect the electric field between plates,capacitance,potential diff between plates,charge stored???????`

jssrk

Sme energy wil b cnvrtd into thermal energy. . . Odr s i hope remains unafctd. .

tarinbansal

Is it a capacitor anymore?? I dont think so.

A capacitor shud have an insulator in between, but U connected the 2 conducting plates with a conducting wire and a resistance, it means its simply a resistive circuit.

rahilarora

yeah i think tarin is right

it will become a simple resistive circuit as the complete charge stored b/w plates will flow through the resistive wire and some will be lost in the form of heat energy

by the way thanks for your rating 2-3 days ago although my answers were incorrect

jssrk

it will b a ressistive circuit nw.Sme energy wil b cnvrtd into thermal energy.

kaymant

With the resistor between the capacitor plates, it will be a RC circuit now, and all equations of RC circuits are applicable.. the field between the plates will increase gradually and reach the maximum value when the capacitor is fully charged.

tarinbansal

RC circuit is when a resistance and a capacitor are there in a circuit either in a series or a parellel connection.

But here, in this case, if U have connected 2 conducting plates with a wire, we dont have a capacitor anymore. It shud be a pure resistive circuit now.

animal

the potential of the 2 plates will be different initially

so a current will flow from higher potential to lower until a state of eqlm is reached when no current flows through the resistor.

when the potential of both the plates will be same then there will be no electric field as there is no electric field for equipotential surface.

tell me if i m correct????????

mukulaish

as the plates are connected by the conductor the charge will flow til the potential of the plates will become equal...
now the total charge on the plates =q-q=0

therefore there would not be any electric field, potential,m charge and hence capacitance.....

smarsh

WE CANT SAY ANYTHING ABT THE CHARGE STORED BECAUSE ITS NO LONGER THE CAPACITOR AS STATED BY @TARIN ......THE ELECTRIC FIELD BETWEEN THE TWO PLATES WILL BECOME ZERO AND THE POTENTIAL DIFFERENCE BETWEEN THE TWO PLATES WILL BE O (SINCE THE CHARGE DISTRIBUTION WILL TAKE PLACE AS LONG AS THE CHARGES ON TWO PLATES BCM SAME...., AND SO DOES THE POTENTIAL)

kaymant

@tarinbansal

Do you see any difference between the two circuits?

tarinbansal

Yes obviously there is a difference.

In second circuit, the capacitor and resistor are seperate.

But in the first circuit, there is no capacitor.

See, the thing is that capacitor plates are conducting. If they were non-conducting, U cud have said that first circuit is a vaild RC circuit.
But the plates are conducting and U join 2 conducting plates with a wire, then the +ve and -ve charges wud flow thru the wire and it will become a pure resistive circuit.

animal

when i told my view on the ques which acc to me is right then y r u going mad abt that it is a rc circuit or not

which eqn u want to apply even if it is a rc circuit ?

but i think that it is not a rc circuit.

tell me if i m correct???????????????

kaymant

@tarinbansal

First of all, we cannot a capacitor made from non-conductors (though the capacitance is still there).

You said the charge will flow down the resistor and therefore there wont be a capacitor. I understand what you meant to say is that since the two plates are connected via a resistor, all the charges will flow and finally there wont be any charge on them and so there will be no capacitor.

There is a problem with that view. You see since the plates are connected to the two ends of the battery, the potential difference between them should be same as the battery. That is the potential difference between the two plates is fixed at that of the battery (by the way this same potential difference exists between the two ends of the resistance in both the circuits). But between two parallel plates if there is a potential difference it, there must be charge on the plates ( I am not saying that if some object is having some potential, it must have some charge). And so there WILL be charges on the two plates. (And so the capacitor will be there.) The same situation in both the circuits: in terms of voltage across each elements, the fields, the current and so the two circuits are exactly the same.

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The plates of capacitor,fully charged are conected through a resistor.Will its presence affect the electric field between plates,capacitance,potential diff between plates,charge stored???????

`The plates of capacitor,fully charged are conected through a resistor.Will its presence affect the electric field between plates,capacitance,potential diff between plates,charge stored???????`