An airfilled parallelplate capacitor has plates of area 260

An air-filled parallel-plate capacitor has plates of area 2.60 cm² separated by 2.00 mm. The capacitor is connected to a 6.0-V battery.

(a) Find the value of its capacitance.
pF

(b) What is the charge on the capacitor?
pC

(c) What is the magnitude of the uniform electric field between the plates?
V/m

Solution

(a)   given that A=2.60*10^(-4) m

d= 2*10^(-3) m

V= 6 v

required formula is

C=(e0*A)/d

where e0 is permittivity of air which is = 8.84*10^(-12)

C=((8.84*10^(-12))*(2.60*10^(-4)))/(2*10^(-3))

C= 1.15*10^(-12) F

C=1.15pF

(b)    we know that Q=C*V

so Q=(1.15*10^(-12))*(6)

Q=6.9*10^(-12) C

Q=6.9 pC

(c)   E=sigma/eo

sigma = Q/A

sigma=(6.9*10^(-12))/(2.6*10^(-4))

sigma= 2.65*10^(-8)

E=(2.65*10^(-8))/(8.84*10^(-12))

E= 2.99*10^3 v/m