Question

Derive the expression for Electrostatic Field Intensity due to a solid spherical mass with uniform charge Q and radius R. Please include the various cases.

#Revision Q9.​

Answer 1

EXPLANATION.

Expression for electrostatic field intensity due to a solid spherical mass with uniform charge Q and radius R.

(1) = Electric field due to charged conducting sphere.

Any charge given to a conductor resides on the surface of conductors,

σ = Surface Charge Density.

σ = Charge/Area.

(1) = Outside at distance r from Center.

r > R.

$\sf \implies \phi_n _e _t = \oint \vec {E}. \vec {dA}$

$\sf \implies \phi_n_e_t = \oint EdA . Cos(0\degree)$

$\sf \implies \phi _n_e_t = \oint EdA$

As we know that,

dA = Area of the sphere = 4πr².

$\sf \implies \phi_n_e_t = E \oint dA$

$\sf \implies \phi _n_e_t = E \ X \ 4 \pi r^{2}$

As we know that,

σ = Charge/Area = Q/4πr².

Q = σ4πr².

$\sf \implies \phi _i_n_s_i_d_e = \dfrac{Q _i_n}{\epsilon_0}$

$\sf \implies \dfrac{Q_i_n}{\epsilon_0} = \dfrac{\sigma 4\pi r^{2} }{\epsilon _0}$

$\sf \implies E \ X \ 4\pi r^{2} = \dfrac{\sigma 4\pi r^{2} }{\epsilon_0}$

$\sf \implies E = \dfrac{\epsilon r^{2} }{\epsilon_0 r^{2} } \ \ or \ E\propto \dfrac{1}{r^{2} }$

$\sf \implies \phi_n_e_t = E \ X \ 4\pi r^{2}$

$\sf \implies \phi_n_e_t = \dfrac{Q_i_n}{\epsilon_0} = \dfrac{Q}{\epsilon_0}$

$\sf \implies E \ X \ 4\pi r^{2} = \dfrac{Q}{\epsilon_0}$

$\sf \implies \boxed{E = \dfrac{Q}{4\pi \epsilon_0 r^{2} } }$

(2) = Inside at distance r from Center.

r < R.

$\sf \implies \phi_n_e_t = \oint \vec {E}. \vec {dA}$

$\sf \implies \phi_n_e_t = \oint \vec {E} . \vec {dA}.Cos(180\degree)$

$\sf \implies \phi_n_e_t = \oint E.dA(-1)$

As we know that,

dA = Total Surface Area = 4πr².

$\sf \implies \phi_n_e_t = -E\oint dA$

$\sf \implies -E \ X \ 4\pi r^{2}$

$\sf \implies \phi_n_e_t = \dfrac{Q_i_n}{\epsilon_o}$

$\sf \implies \phi_n_e_t = 0$

$\sf \implies \phi_i_n = 0$

$\sf \implies -E \ X \ 4\pi r^{2} = 0$

$\sf \implies \boxed{E = 0}$

Under Equilibrium E field inside a conductor is always zero.

(3) = E vs r.

$\sf \implies inside = Q_i_n = 0 \ and \ E = 0$

$\sf \implies outside = E = \dfrac{Q}{4\pi r^{2} \epsilon_o} = \dfrac{\sigma R^{2} }{\epsilon_o r^{2} } = E \propto \dfrac{1}{r^{2} }$

$\sf \implies E_m_a_x = Surface$

$\sf \implies r = R$

$\sf \implies E = \dfrac{\sigma R^{2} }{\epsilon _oR^{2} }$

$\sf \implies E = \dfrac{\sigma}{\epsilon_o}$

(2) = Electric Field due to Non-Conducting (insulator) Sphere.

Any charge given to Non-Conductor/dielectric/insulator remains at its place.

ρ = Volume Charge Density.

ρ = Charge/Volume.

(1) = Outside at distance r from Center.

r > R.

$\sf \implies \phi_n_e_t = E \ X \ 4\pi r^{2}$

$\sf \implies \rho = \dfrac{Q}{\dfrac{4}{3} \pi R^{3} }$

$\sf \implies Q = \rho \ X \ \dfrac{4}{3} \pi R^{3}$

$\sf \implies \phi_n_e_t = \dfrac{Q_i_n}{\epsilon_o} = \dfrac{\rho \ X \ \dfrac{4}{3} \pi R^{3} }{\epsilon_o}$

$\sf \implies E \ X \ 4\pi r^{2} = \dfrac{\rho \ X \ \dfrac{4}{3}\pi R^{3} }{\epsilon_o}$

$\sf \implies E = \dfrac{\rho R^{3} }{3\epsilon_o r^{2} }$

Let total charge on sphere be Q.

$\sf \implies E \ X \ 4\PI r^{2} = \dfrac{Q_i_n}{\epsilon_o} = \dfrac{Q}{\epsilon_o}$

$\sf \implies \boxed{E = \dfrac{Q}{4\pi r^{2}\epsilon_o } \ and \ E \propto \dfrac{1}{r^{2} }}$

(2) = inside at distance r from centre.

r < R.

$\sf \implies \phi_n_e_t = \oint E.dA$

$\sf \implies \phi_n_e_t = \oint E.dA Cos(0\degree)$

$\sf \implies \phi_n_e_t = E \oint dA$

$\sf \implies \rho = \dfrac{Q}{\dfrac{4}{3} \pi r^3} }$

$\sf \implies Q = \rho \ X \ \dfrac{4}{3}\pi r^{3}$

$\sf \implies \phi_n_e_t =\dfrac{Q_i_n}{\epsilon_o} = \dfrac{\rho \ X \ \dfrac{4}{3}\pi r^{3} }{\epsilon_o}$

$\sf \implies E \ X \ 4\pi r^{2} = \dfrac{\rho \ X \ \dfrac{4}{3}\pi r^{3} }{\epsilon_o}$

$\sf \implies E = \dfrac{\rho r^{3} }{3\epsilon_o r^{2} }$

$\sf \implies E = \dfrac{\rho r}{3\epsilon_o}$

Inside E ≠ 0 Non-conducting insulating sphere.

E ∝ r.

At center

r = 0

E = 0

Let total charge on surface be Q.

$\sf \implies E = \dfrac{\rho r}{3\epsilon_o}$

$\sf \implies E = \dfrac{Q}{\dfrac{4}{3}\pi R^{3} } \ X \ \dfrac{r}{3\epsilon_o}$

$\sf \implies \boxed{E = \dfrac{Qr}{4\pi \epsilon_o R^{3} } }$

(3) = E vs r.

$\sf \implies E = \dfrac{\rho r}{3\epsilon_o}$

$\sf \implies inside = E = \dfrac{Qr}{4\pi \epsilon_o R^{3} } = E \propto r$

$\sf \implies outside = E = \dfrac{Q}{4\pi \epsilon_o r^{2} } = E \propto \dfrac{1}{r^{2} }$

$\sf \implies E_m_a_x = Surface$

$\sf \implies r = R$

$\sf \implies \boxed{E_s_u_r_f_a_c_e = \dfrac{Q}{4\pi \epsilon_oR^{2} } }$

Answer 2

Answer:

hope its correct...

considering it has non conducting...solid sphere

If we take conducting then everything will be same beside inside electric field...as all charge will go on the surface for equilibrium and inside E=0 in the conducting solid sphere