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Energy stored in parallel plate capacitor

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  1. Mar 14, 2015 #1
    Can anyone please tell me whether I have solved this question correctly or not.

    1. The problem statement, all variables and given/known data

    A parallel plate capacitor has a charge of 0.040µC on each plate with a potential difference of 240 V. The parallel plates are separated by 0.20 mm of air. What energy is stored in this capacitor?

    2. Relevant equations
    Q/V=eo A/d
    Cair=Aeo/d
    E=1/2 CV2
    3. The attempt at a solution
    Q/V=eo A/d

    A=Qd/Veo

    A=(0.040u)(0.20/1000)/(240)(8.85x10-12)

    A=3.766x10-3 m2

    Cair=Aeo/d

    =[3.766x10-3(8.85x10-12)]/(0.20/1000)

    =0.1666uF

    E=1/2 CV2

    =1/2(0.1666u)(240)2

    =4.7994uJ
     
  2. jcsd
  3. Mar 14, 2015 #2

    gneill

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    Staff: Mentor

    You have reached the correct result. But you took a pretty long route to get there. Note that the relationship between charge and potential on a capacitor is Q = C*V. You've been given both Q and V so you could have obtained C directly.
     
  4. Mar 14, 2015 #3
    Alright thankyou!!
     
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