The electric field intensity due to surface charge distribution is given by

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BPSC AE Paper 5 (Electrical) 25 Mar 2022 Official Paper

\(\int {\frac{{{\rho _l}dl}}{{4\pi {\varepsilon _0}|r{|^2}}}} {\hat a_r}\)
\(\int {\frac{{{\rho _s}ds}}{{4\pi {\varepsilon _0}|r{|^2}}}} {\hat a_r}\)
\(\int\int {\frac{{{\rho _s}ds}}{{4\pi {\varepsilon _0}|r{|^2}}}} {\hat a_r}\)
\(\int\int\int {\frac{{{\rho _v}dv}}{{4\pi {\varepsilon _0}|r{|^2}}}} {\hat a_r}\)

Answer 1

Concept:

The electric field intensity due to a point charge is given by:

\(E = {1 \over 4\pi \epsilon_o}{Q\over |r|^2}{ \hat a_r}\) ...........(i)

If the point charges are uniformly distributed over a surface, then the electric field intensity is:

\(Q=\iint{}^{}\rho_s\space ds\) ............(ii)

Putting the value of equation (ii) in equation (i), we get:

\(E=\int\int {\frac{{{\rho_s}ds}}{{4\pi {\varepsilon_0}|r{|^2}}}} {\hat a_r}\)

E = Electric field intensity

ρs = Surface charge density

âr = Unit surface vector

Additional Information The electric field intensity due to line charge distribution is given by:

\(E=\int {\frac{{{ρ_l}dl}}{{4\pi {\varepsilon_0}|r{|^2}}}} {\hat a_r}\)

ρl = Line charge Density

The electric field intensity due to volume charge distribution is given by:

\(E=\int \int \int {\frac{{{\rho_v}dv}}{{4\pi {\varepsilon_0}|r{|^2}}}} {\hat a_r}\)

ρv = Volume charge Density