Asked by anonymous
can anyone help me on this question. I don't understand
A capacitor is constructed of two aluminium plates separated by a dielectric. The capacitance is measured to be 268.00μF. Without changing the dimensions of the plates, they are brought closer together so that the distance between the two plates is 1/2 of the original distance. What is the new capacitance (in μF)?
A capacitor is constructed of two aluminium plates separated by a dielectric. The capacitance is measured to be 268.00μF. Without changing the dimensions of the plates, they are brought closer together so that the distance between the two plates is 1/2 of the original distance. What is the new capacitance (in μF)?
Answers
Answered by
bobpursley
C= ke<sub>o</sub> * Area/ distance apart
wouldn't it double?
wouldn't it double?
Answered by
anonymous
i don't know if it would double because this is all the information i got from my professor.
Answered by
anonymous
I checked the solutions and it says 536.00 but i don't know where 536.00 came from. Can someone help me!
Answered by
Damon
2 * 268 = 536
Answered by
Damon
Bob gave you the formula
C= k eo * Area/ distance apart
divide distance by 2 ----> multiply C by 2
C= k eo * Area/ distance apart
divide distance by 2 ----> multiply C by 2
Answered by
bobpursley
C= keo * Area/ distance apart If distance apart changes to 1/2 original, then if
Corig=keo*area/d<sub>o</sub>=268 then
Cnew=Corig /1/2 = 2 Corig
Corig=keo*area/d<sub>o</sub>=268 then
Cnew=Corig /1/2 = 2 Corig
Answered by
anonymous
ok thanks
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