Question

A spherical 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?

Answer 1

Answer: 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=-$\int\limits^a_b {E} \, dx$

Where E = k*Q/ r^2

so we have $\int\limits^a_b {K+Q1/r} \, dr$

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

Answer 2

(a) The capcitance is 73.7 pF

(b) Electric field E₁ is 4.86×10³ N/C

(c) Electric field E₂ is 3.68×10³ N/C

Given that the radius of the inner sphere is $r_a=12.3cm$ and the radius of the outer sphere is $r_b=15.1cm$, separated by a vacuum.

Capacitance

(a)The capacitance is given by:

$C=\frac{4\pi \epsilon_o ab}{b-a}\\\\C=\frac{12.3\times10^{-2}\times15.1\times10^{-2}}{9\times10^9(2.8\times10^{-2})}F\\\\C=73.7 \;pF$

The charge Q on the capacitor:

Q = CV

$Q=73.7\times10^{-12}\times120\\\\Q=8.844\times10^{-9}C$on each sphere.

Electric field

The charge on the outer spherical shell of the capacitor does not contribute to the electric field inside the capacitor.

(b) The electric field at a point r = 12.8cm is given by:

$E_1=k\frac{Q}{r^2}$

where k is the Coulomb's constant.

$E_1=9\times10^9\frac{8.844\times10^{-9}}{(12.8)^2\times 10^{-4}}N/C\\\\E_1=4.86\times10^3N/C$

(c) The electric field at a point r = 14.7cm is:

$E_2=9\times10^9\frac{8.844\times10^{-9}}{(14.7)^2\times 10^{-4}}N/C\\\\E_2=3.68\times10^3N/C$

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