A parallel plate capacitor is charged to 250 pC (that is, one plate has charge +

Question

A parallel plate capacitor is charged to 250 pC (that is, one plate has charge +250 pC and the other plate has charge –250 pC). The potential difference between the plates is 45.0V.  

a) If the area of each plate is 6.4 cm2 , what is the plate separation? What is the electric field strength between the plates?

b) Assume the plate separation doubles from the value calculated in (a). What must the potential across the plates be for the charge on the plates to remain at 250 pC? What is the electric field strength between the plates?

show step by step

a) d = 1.02x10^-3 m

E = 4.41x10^4 V/m

b) Vnew = 2V

Vold = 90V

A)

Explanation / Answer

here Q = 250 pC = 250 x 10-12 C A = 6.4 cm2 = 6.4 x 10-4 .m2   V = 45 volt

(a) so the plate seperation

C = eoA / d

here C = Q / V = (250 x 10-12) / 45 = 5.5555 x 10-12 F

so d = (8.85 x 10-12) (6.4 x 10-4) / (5.5555 x 10-12)

d = 10.1952 x 10-4 m Ans

so the electric field strength between the plates

E = V / d

= 45 / (10.1952 x 10-4) = 4.4138 x 104 V/m Ans

(b) if the plate seperation is double then

d = 2 x 10.1952 x 10-4 = 20.3904 x 10-4 m

then the capacitance

C = eoA / d

= (8.85 x 10-12) (6.4 x 10-4) / (20.3904 x 10-4)

C = 2.7777 x 10-12 F

as we know that C = Q / V

then V = Q / C

V = (250 x 10-12) / (2.7777 x 10-12)   

V = 90 volt Ans

so the electric field

E = V / d

E = 90 / (20.3904 x 10-4)

E = 4.41384 x 104 V/m Ans

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