Side view Top view a) A capacitor is constructed of identical parallel plates ea

Question

Side view Top view a) A capacitor is constructed of identical parallel plates each of area A 5 m2 that overlap only halfway as shown above. The spacing is d= 5 cm. Neglect the fields at the edges of the plates. Suppose a charge of Q= 4-10-8 C is placed on the capacitor. The voltage is then V1 ; then the plates are moved without changing the spacing, so that now they overlap completely, and the voltage across them is now measured to be V2. The values Vi and V2 are thern b) The plates of the capacitor in the above question are now connected to a Vattery-17Volt battery. The plates are initially separated, with 50% overlap, as shown in the Figure above. when the plates are moved (parallel to each other with the spacing between them kept constant and still connected to the batter) so that their overlap is 100% Charge is forced from the plates back into the battery, leaving Q 3.0x108 C on the capacitor. The total charge on the capacitor remains the same, Q -1.5x10-8 C Charge is forced from the plates back into the battery, leaving Q = 0.8×10-8 C on the capacitor O Charge is drawn from the battery, leaving Q 3.0x10-8 Con the capacitor. Charge is drawn from the battery, leaving Q 1.5x108 C on the capacitor The potential across the capacitor drops to 8.5V

Explanation / Answer

Firstly we will gind the capacitance in both the cases.

C=€°A/d where €°=constant=8.85*10^—5

A=area OVERLAPPING

d=distance or separation between plates.

So when plates are half overlapped

Capacitance C1=€°A/d

A=2.5m² and d=5cm=0.5m

C1=8.85*10^-12*2.5*/0.05

C1=4.425*10—^10 farad.

Now Q=CV where q=charge=4*10^-8, C is capacitance and V is voltage.

Put values

4*10^-8=4.425*10^-10*V

V1=90.45 volts.

When capacitors are fully overlapped,

Capacitance =8.85*10^-12* 5* /5*10^-2

Capacitance =8.85*10^-10 farad.

Also q=cv

4*10^-8=8.85*10^-10*V2

V2=45.197 volts.

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