Assuming that the charge 'q' is uniformly distributed in a spherical volume of radius 'R'. Discuss the variation of a) electric intensity b) potential as the field point is moved from the center of the sphere to infinity.
Solution.
a)
We conclude our spherical volume of a symmetrical surface S S S r > R r > R r > R S S S
Φ E = ∯ S E d S = E 4 π r 2 ; \Phi_ {E} = \oiint_ {S} E d S = E 4 \pi r ^ {2}; Φ E = ∬ S E d S = E 4 π r 2 ;
By Gauss's law:
Φ E = q ε 0 ; \Phi_ {E} = \frac {q}{\varepsilon_ {0}}; Φ E = ε 0 q ;
Therefore:
E 4 π r 2 = q ε 0 ; E 4 \pi r ^ {2} = \frac {q}{\varepsilon_ {0}}; E 4 π r 2 = ε 0 q ;
The electric field intensity outside the spherical volume r > R r > R r > R
E = q 4 π r 2 ε 0 . E = \frac {q}{4 \pi r ^ {2} \varepsilon_ {0}}. E = 4 π r 2 ε 0 q .
At the point B ′ B' B ′ r < R r < R r < R q ′ q' q ′ r ′ r' r ′ S ′ S' S ′
Φ E = ∯ S ′ E d S = E 4 π r ′ 2 ; \Phi_ {E} = \oiint_ {S ^ {\prime}} E d S = E 4 \pi r ^ {\prime 2}; Φ E = ∬ S ′ E d S = E 4 π r ′2 ;
By Gauss's law:
Φ E = q ′ ε 0 ; \Phi_{E} = \frac{q'}{\varepsilon_{0}}; Φ E = ε 0 q ′ ; q ′ = ρ q V ′ ; q' = \rho_{q} V'; q ′ = ρ q V ′ ;
The volume charge density:
ρ q = q V ; \rho_{q} = \frac{q}{V}; ρ q = V q ; V = 4 3 π R 3 ; V = \frac{4}{3} \pi R^{3}; V = 3 4 π R 3 ; ρ q = 3 q 4 π R 3 . \rho_{q} = \frac{3q}{4\pi R^{3}}. ρ q = 4 π R 3 3 q . q ′ = 3 q 4 π R 3 V ′ ; q' = \frac{3q}{4\pi R^{3}} V'; q ′ = 4 π R 3 3 q V ′ ; V ′ = 4 3 π r ′ 3 ; V' = \frac{4}{3} \pi r'^3; V ′ = 3 4 π r ′3 ; q ′ = 3 q 4 π R 3 ⋅ 4 3 π r ′ 3 = q r ′ 3 R 3 . q' = \frac{3q}{4\pi R^{3}} \cdot \frac{4}{3} \pi r'^3 = \frac{q r'^3}{R^{3}}. q ′ = 4 π R 3 3 q ⋅ 3 4 π r ′3 = R 3 q r ′3 . Φ E = q r ′ 3 ε 0 R 3 . \Phi_{E} = \frac{q r'^3}{\varepsilon_{0} R^{3}}. Φ E = ε 0 R 3 q r ′3 . E 4 π r ′ 2 = q r ′ 3 ε 0 R 3 ; E 4\pi r'^2 = \frac{q r'^3}{\varepsilon_{0} R^{3}}; E 4 π r ′2 = ε 0 R 3 q r ′3 ; E = q r ′ 4 π ε 0 R 3 . E = \frac{q r'}{4\pi \varepsilon_{0} R^{3}}. E = 4 π ε 0 R 3 q r ′ .
The electric field intensity inside the spherical volume r < R r < R r < R
E = q r 4 π ε 0 R 3 . E = \frac{q r}{4\pi \varepsilon_{0} R^{3}}. E = 4 π ε 0 R 3 q r .
b)
The electric potential outside the spherical volume r > R r > R r > R
d V E = − E d r = q 4 π ε 0 r 2 d r ; d V_{E} = - E d r = \frac{q}{4\pi \varepsilon_{0} r^{2}} d r; d V E = − E d r = 4 π ε 0 r 2 q d r ; V E = − E ∫ ∞ r d r = − q 4 π ε 0 ∫ ∞ r d r r 2 ; V_{E} = - E \int_{\infty}^{r} d r = - \frac{q}{4\pi \varepsilon_{0}} \int_{\infty}^{r} \frac{d r}{r^{2}}; V E = − E ∫ ∞ r d r = − 4 π ε 0 q ∫ ∞ r r 2 d r ; V E = q 4 π ε 0 ( 1 r − 1 ∞ ) . V _ {E} = \frac {q}{4 \pi \varepsilon_ {0}} \Big (\frac {1}{r} - \frac {1}{\infty} \Big). V E = 4 π ε 0 q ( r 1 − ∞ 1 ) . V E = q 4 π ε 0 r . V _ {E} = \frac {q}{4 \pi \varepsilon_ {0} r}. V E = 4 π ε 0 r q .
The electric potential inside the spherical volume r < R r < R r < R
V E = − E ∫ ∞ r d r = − ∫ ∞ r q 4 π ε 0 r 2 d r − ∫ R r q 4 π ε 0 R 3 r d r ; V _ {E} = - E \int_ {\infty} ^ {r} d r = - \int_ {\infty} ^ {r} \frac {q}{4 \pi \varepsilon_ {0} r ^ {2}} d r - \int_ {R} ^ {r} \frac {q}{4 \pi \varepsilon_ {0} R ^ {3}} r d r; V E = − E ∫ ∞ r d r = − ∫ ∞ r 4 π ε 0 r 2 q d r − ∫ R r 4 π ε 0 R 3 q r d r ; − ∫ R r q 4 π ε 0 R 3 r d r = − q 4 π ε 0 R 3 ⋅ r 2 2 + q 4 π ε 0 R 3 ⋅ R 2 2 ; - \int_ {R} ^ {r} \frac {q}{4 \pi \varepsilon_ {0} R ^ {3}} r d r = - \frac {q}{4 \pi \varepsilon_ {0} R ^ {3}} \cdot \frac {r ^ {2}}{2} + \frac {q}{4 \pi \varepsilon_ {0} R ^ {3}} \cdot \frac {R ^ {2}}{2}; − ∫ R r 4 π ε 0 R 3 q r d r = − 4 π ε 0 R 3 q ⋅ 2 r 2 + 4 π ε 0 R 3 q ⋅ 2 R 2 ; V E = 3 q 8 π ε 0 R − q r 2 8 π ε 0 R 3 . V _ {E} = \frac {3 q}{8 \pi \varepsilon_ {0} R} - \frac {q r ^ {2}}{8 \pi \varepsilon_ {0} R ^ {3}}. V E = 8 π ε 0 R 3 q − 8 π ε 0 R 3 q r 2 .
Answer:
a)
The electric field intensity inside the spherical volume r < R r < R r < R
E = q r 4 π ε 0 R 3 . E = \frac {q r}{4 \pi \varepsilon_ {0} R ^ {3}}. E = 4 π ε 0 R 3 q r .
The electric field intensity outside the spherical volume r > R r > R r > R
E = q 4 π r 2 ε 0 . E = \frac {q}{4 \pi r ^ {2} \varepsilon_ {0}}. E = 4 π r 2 ε 0 q .
b)
The electric potential inside the spherical volume r < R r < R r < R
V E = 3 q 8 π ε 0 R − q r 2 8 π ε 0 R 3 . V _ {E} = \frac {3 q}{8 \pi \varepsilon_ {0} R} - \frac {q r ^ {2}}{8 \pi \varepsilon_ {0} R ^ {3}}. V E = 8 π ε 0 R 3 q − 8 π ε 0 R 3 q r 2 .
The electric potential outside the spherical volume r > R r > R r > R
V E = q 4 π ε 0 r . V _ {E} = \frac {q}{4 \pi \varepsilon_ {0} r}. V E = 4 π ε 0 r q . 
