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Physics: Post your doubts here!

Q

qdkhan

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amal sharkawi said:
View attachment 54032

THIS ONE ALSO PLZ ......
Click to expand...
In the circuit shown, let the current be I1
I1=V/R=6/2R
Power dissipated in external resistor (P1)=I1*I1*resistance of external resisitor=(6/2R)*(6/2R)*R= 36R/(4R*R)=9/R
in the circuit with negligible internal resistance, let the current be I2
I2=V/R=6/R
so power dissipated in xternal circuit(P2)=I2*I2*Resistance of external resistor=(6/R)*(6/R)*R=36R/(R*R)=36/R
SO P2=4P1
 
princess Anu

princess Anu

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qwertypoiu said:
When the negatively charged plate is brought close to the initially neutral plate, the electrons in the bottom plate are repelled away and move into the earth, since a wire is connected so the electrons are able to flow away. This makes the bottom plate become overall positively charged, since there are more protons than electrons.
Click to expand...
:confused:
but isnt it that anything earthed becomes neutral? :eek: :unsure:
do u mean if the bottom plate was not earthed, it wud not get a +ve charge?
 
qwertypoiu

qwertypoiu

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princess Anu said:
:confused:
but isnt it that anything earthed becomes neutral? :eek: :unsure:
do u mean if the bottom plate was not earthed, it wud not get a +ve charge?
Click to expand...
If the bottom plate was not earthed, there will be positive charge in upper part of this plate and negative charge in bottom half, because the elections are repelled and so move down. They are not able to escape anywhere. So top part would be positive and bottom part would be negative, within the bottom plate itself.

By earthing it we allow electrons to move away so the plate is positively charged everywhere as a whole
 
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