Question #28594

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?

Expert's answer

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 SS of radius r>Rr > R . Electric flux through the surface SS is equal to:


ΦE=SEdS=E4πr2;\Phi_ {E} = \oiint_ {S} E d S = E 4 \pi r ^ {2};


By Gauss's law:


ΦE=qε0;\Phi_ {E} = \frac {q}{\varepsilon_ {0}};


Therefore:


E4πr2=qε0;E 4 \pi r ^ {2} = \frac {q}{\varepsilon_ {0}};


The electric field intensity outside the spherical volume r>Rr > R :


E=q4πr2ε0.E = \frac {q}{4 \pi r ^ {2} \varepsilon_ {0}}.


At the point BB' , inside of the spherical volume r<Rr < R electric field is determined by only a charge qq' inside the sphere of a radius of rr' . Electric flux through the surface SS' is equal to:


ΦE=SEdS=E4πr2;\Phi_ {E} = \oiint_ {S ^ {\prime}} E d S = E 4 \pi r ^ {\prime 2};


By Gauss's law:


ΦE=qε0;\Phi_{E} = \frac{q'}{\varepsilon_{0}};q=ρqV;q' = \rho_{q} V';


The volume charge density:


ρq=qV;\rho_{q} = \frac{q}{V};V=43πR3;V = \frac{4}{3} \pi R^{3};ρq=3q4πR3.\rho_{q} = \frac{3q}{4\pi R^{3}}.q=3q4πR3V;q' = \frac{3q}{4\pi R^{3}} V';V=43πr3;V' = \frac{4}{3} \pi r'^3;q=3q4πR343πr3=qr3R3.q' = \frac{3q}{4\pi R^{3}} \cdot \frac{4}{3} \pi r'^3 = \frac{q r'^3}{R^{3}}.ΦE=qr3ε0R3.\Phi_{E} = \frac{q r'^3}{\varepsilon_{0} R^{3}}.E4πr2=qr3ε0R3;E 4\pi r'^2 = \frac{q r'^3}{\varepsilon_{0} R^{3}};E=qr4πε0R3.E = \frac{q r'}{4\pi \varepsilon_{0} R^{3}}.


The electric field intensity inside the spherical volume r<Rr < R:


E=qr4πε0R3.E = \frac{q r}{4\pi \varepsilon_{0} R^{3}}.


b)

The electric potential outside the spherical volume r>Rr > R:


dVE=Edr=q4πε0r2dr;d V_{E} = - E d r = \frac{q}{4\pi \varepsilon_{0} r^{2}} d r;VE=Erdr=q4πε0rdrr2;V_{E} = - E \int_{\infty}^{r} d r = - \frac{q}{4\pi \varepsilon_{0}} \int_{\infty}^{r} \frac{d r}{r^{2}};VE=q4πε0(1r1).V _ {E} = \frac {q}{4 \pi \varepsilon_ {0}} \Big (\frac {1}{r} - \frac {1}{\infty} \Big).VE=q4πε0r.V _ {E} = \frac {q}{4 \pi \varepsilon_ {0} r}.


The electric potential inside the spherical volume r<Rr < R :


VE=Erdr=rq4πε0r2drRrq4πε0R3rdr;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;Rrq4πε0R3rdr=q4πε0R3r22+q4πε0R3R22;- \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};VE=3q8πε0Rqr28πε0R3.V _ {E} = \frac {3 q}{8 \pi \varepsilon_ {0} R} - \frac {q r ^ {2}}{8 \pi \varepsilon_ {0} R ^ {3}}.


Answer:

a)

The electric field intensity inside the spherical volume r<Rr < R :


E=qr4πε0R3.E = \frac {q r}{4 \pi \varepsilon_ {0} R ^ {3}}.


The electric field intensity outside the spherical volume r>Rr > R :


E=q4πr2ε0.E = \frac {q}{4 \pi r ^ {2} \varepsilon_ {0}}.


b)

The electric potential inside the spherical volume r<Rr < R :


VE=3q8πε0Rqr28πε0R3.V _ {E} = \frac {3 q}{8 \pi \varepsilon_ {0} R} - \frac {q r ^ {2}}{8 \pi \varepsilon_ {0} R ^ {3}}.


The electric potential outside the spherical volume r>Rr > R :


VE=q4πε0r.V _ {E} = \frac {q}{4 \pi \varepsilon_ {0} r}.

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