Consider the following equation : c=q/v Define voltage as it relates to capacita

Question

Consider the following equation : c=q/v

Define voltage as it relates to capacitance.

Is voltage the amount of work needed to keep charge on the capactior? In a circuit with just a capactior and a battery, why is it when you disconnect the battery voltage increases but the charge Q stays the same? Does the capacitence decrease in this scenario?

I'm really having a hard time understanding these concepts. Please be as detailed as possible when connecting concepts.

The question below is why I am asking the question above. I got the question right I just don't know why I got it right and the difference between battery connected to a capacitor versus a battery disconnected from a capacitor.

A parallel-plate capacitor has capacitance 4.50 F (a) How much energy is stored in the capacitor if it is connected to a 8.00-V battery? (b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? (c) The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored?

Explanation / Answer

a)

Energy stored in capacitor is given by

E=(1/2)CV2=(1/2)*4.5*82

E=144 uJ

b)

Q=CV= 4.5*8 =36 uC

If the battery is disconnected ,the charge remians constant

C=eoA/d

=>Cnew=(d/dnew)C =(1/2)*4.5 =2.25 uF

Energy stored

E=(1/2)(Q2/C) =(1/2)(362/2.25) =288 uJ

c)

E=(1/2)CV2=(1/2)*2.25*82

E=72 uJ

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