SUNG bors and Capacitan Parallel Plate Capacitor and Battery a Pi Capcitor m Two

Question

SUNG bors and Capacitan Parallel Plate Capacitor and Battery a Pi Capcitor m Two parallel plates, each having area A . 3687 cm2 are connected to the terminats of a battery of voltage V 6 V as shown The plates are separated by a distance d 0.58 What is Q, the charge on the top plate? 2 What is LI the energy stored in the this capaci tor? s the this ca 3) The battery is now discannected from the plates and the separation of the plates is doubled (- 1.16 cm) What is the energy stored in this new capacitor? 4) What is E, the magnitude of the electric field in the region between the plates? O ) Compare V, the magnitude of the new potential difference across the plates, to Vi, the voltage of the bettery 2d Two uncharged paraliel plates are now connected to the initial pair of plates as shown. How will the electric field, E, and potential difference across the plates·V, change, if at all? Both E and V will reman the sae GE will decrease and V wil increase E will ncrease and V will decrease Both E and V will decresse O Both E and V will increase

Explanation / Answer

a)

Capacitane

C=eoA/d =(8.8542*10-12)*(3687*10-4)/(0.58*10-2)=5.63*10-10 F

charge

Q=CV =6*(5.63*10-10)=3.377*10-9C

b)

Energy stored in capacitor is

U=(1/2)CV2=(1/2)*(5.63*10-10)*62=1.013*10-8J

c)

Capacitance

C=(8.8542*10-12)(3687*10-4)/(1.16*10-2)=2.81*10-10 F

Since battery is disconnected charge remains constant.Therefore voltage across capacitor

V=Q/C =(3.377*10-9)/(2.81*10-10)=12 Volts

Energy stored

U=(1/2)*(2.81*10-10)*122=2.026*10-8J

d)

E=V/d =12/(1.16*10-2)=1034.5 N/C

e)

V>Vb

f)

Both E and V will decrease

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