A non-conducting sphere of radius R = 5 cm has its center at the origin O of the coordinates system as shown in the figure. It has two spherical cavities of radius r = 1 cm, whose centers are at (0, 3 cm), (0, - 3 cm), respectively, and the solid material of the sphere has a uniform positive charge density \(\rho=\frac{1}{\pi} \mu \mathrm{Cm}^{-3}\).Compute electric potential at point P (4 cm, 0)

F3 Savita Engineering 28-6-22 D1

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  1. 23.68 V
  2. 42.52 V
  3. 65.21 V
  4. 34.92 V

Answer (Detailed Solution Below)

Option 4 : 34.92 V
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Concept 

The electric potential inside a uniformly charged sphere is,

\(V = \frac{kQ}{2R}(3- \frac{r^2}{R^2})\)

Where  \(k = \frac{1}{4π \epsilon_0} = 9× 10^9\)

R = radius of the sphere, r is the distance from the center of the sphere (r < R)

Electric potential outside a uniformly charged sphere is, when ( r > R)

\(V = \frac{q}{4π \epsilon_0 r}\)

Calculation:
Given: Radius of sphere R = 5 cm = 0.05 m

The radius of the cavity r = 0.01m

Charged density ρ = 1/π μ C-m-3

Calculating the charge on the sphere

\(Q = \frac{4\pi}{3}R^3\rho \) = \(\frac{4\pi}{3}(0.05^3)× \frac{1}{\pi}\)

Q = 1.667×10-10C

Charge in the cavity q

\(q = \frac{4\pi}{3}(0.01^3)\frac{1}{\pi}\) = 1.333 ×10-12C

The electric potential at point P (4 cm, 0),

\(V_p = V_{sphere} - 2V_{cavity}\)

\(\Rightarrow V_p = \frac{kQ}{2R}(3-\frac{r^2}{R^2}) - 2\times \frac{kq}{r}\)

\(\Rightarrow V_p = \frac{9\times 10^9\times 1.67\times 10^{-10}}{2\times 0.05}(3-(\frac{4}{5})^2) - 2\times \frac{9\times 10^9 \times 1.667\times 10^{-12}}{0.05}\)

\(\Rightarrow V_p = 9\times 10^9[ 3.94 \times 10^{-9} - 6.668\times 10^{-11}]\)

\(\Rightarrow V_p = 9\times 10^9[ 3.874\times10^{-9}]\)

\(\Rightarrow V_p = 34.92\ V\)

Hence the electric potential inside the charged sphere at the point P is 34.92

Additional Information

Derivation of the electric field inside a charged sphere (r < R)

let \(k = \frac{1}{4\pi \epsilon_0}\)

\(V = - \int\limits_\infty^R \frac{kQ}{r^2}dr -\int\limits_R^r\frac{kQ}{R^3}rdr\)

.\(V = \frac{kQ}{r} - \frac{kQ}{2R^3}(r^2-R^2)\)

\(V= \frac{kQ}{2R}(3-\frac{r^2}{R^2})\)

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