Two 15-cm-diameter electrodes 0.40 cm apart form a parallel-plate capacitor. The

Question

Two 15-cm-diameter electrodes 0.40 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 11 V battery.

(A) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes while the capacitor is attached to the battery? Q=?, E=?, delta V=?

(B) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 0.80 cm apart? The electrodes remain connected to the battery during this process. Q=?, E=?, delta V=?

(C) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after the original electrodes are expanded until they are 30 cm in diameter while remaining connected to the battery? Q=?, E=?, delta V=?

Explanation / Answer

The capacity of a parallel plate capcitor is given by the formula

C = e *A/D where A is the area of the plate and D isthe distance of seperation between the plates

Here D= 0.004 m A= 22/7 * (0.15/2)2 sqm

C= 8.854x10-12 x(22/7)x(0.15/2)2/(0.004) = 0.3913 x 10-10 F

The capacitor is connected toa 11V battery and hence the capacitor will be fully charged to 11V

Part A:

Potential difference between tha plates delta V = 11V

Electric field is the potential difference per unit distance. Here we assume the electric field is uniform inside the capacitor.

Electric field E = 11/0.004 = 2750 V/m

Charge on the plates is Q = CV= 0.3913x10-10 x 11 = 4.3043 x 10-10 Coulombs

Part B :

All parameters remain same only distance changed from 0.4 cm to 0.8 cm

Now D is 0.008 instead of 0.004

Capacity C = 0.3913 x 10-10 /2 = 0.1956 x 10-10 Farads , The capacity will be half of the original as the distance has doubled.

delta V = 11V , the battery is same

Electric filed E = V/d = 11/0.008 = 1375 V/m

Charge on the plates Q= VC = 0.3913 Coulombs, It will be half of original as the capacity has reduced to half

Part C :

The diameter has doubled and the area has increased to 4 times

Hence C = 0.3931 x10-10 x4 = 1.5652 x 10-10 F

delta V remains same =11v as it is connected tothe same battery

Electric filed remains same as the distance between the plates did not change.

E = 11/0.004 = 2750 V/m

Charge Q = V x C , the capacity has increased to 4 times hence the Charge on the plates will be 4 times

                 = 4.3043 x 10-10 x4 = 17.2172 x 10-10 F

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