Aspherical capacitor is formed from two concentric spherical conducting shells separated by vacuum. the inner sphere has a radius of ra = 12.3 cm , and the outer sphere has a radius of rb = 15.1 cm . a potential difference of 120 v is applied to the capacitor.

(a) what is the capacitance of the capacitor?
(b) what is the magnitude e1 of the electric field e? at radius r= 12.8cm, just outside the inner sphere?
(c) what is the magnitude of e? at r= 14.7cm, just inside the outer sphere?

a) 73.41 10^-12 F; b)4.83* 10^3 N/C; c) 3.66 *10^3 N/C

Explanation: To solve this problem we have to consider the following: The Capacity= Charge/Potential Difference

As we know the capacity is  value that depend on the geometry of the capacitor, in our case two concentric spheres.

So Potential Difference between the spheres is given by:

ΔV=-

Where E = k*Q/ r^2

so we have

then

Vb-Va=k*Q(1/b-1/a)=kQ (ab/b-a)

Finally using C=Q/ΔV=ab/(k(b-a))

To caclulate the electric firld we first obtain the charge

Q=ΔV*C=120 V*73.41 10^-12 F=8.8 10^-9 C

so E=KQ/r^2 for both values of r

r=12.8 cm ( in meters)

r2=14.7 cm

E(r1)=4.83* 10^3 N/C

E(r2)=3.66 *10^3 N/C

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.  

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j