Consider a sphere of radius R which carries a uniform charge density . If a sphere of radius iscarved out of it, as shown, the ratio of magnitude of electric field of magnitude of electric field and , respectively, atpoints A and B due to the remaining portion is

JEE MainPhysicsModerateElectrostatics

Consider a sphere of radius R which carries a uniform charge
density $\rho$ . If a sphere of radius $\frac{R}{2}$ iscarved out of it, as
shown, the
ratio $\left| \frac{{\overset{\rightarrow}{E}}_{A}}{{\overset{\rightarrow}{E}}_{B}} \right|$
of magnitude of electric field of magnitude of electric field
${\overset{\rightarrow}{E}}_{A}\$ and ${\overset{\rightarrow}{E}}_{B}$ ,
respectively, atpoints A and B due to the remaining portion is

Show answer and solution

For a solid sphere

E = \frac{\rho r}{3\varepsilon_{0}}

E_{A} = \frac{- \rho R}{2\left( 3\varepsilon_{0} \right)}

\left| E_{A} \right| = \frac{\rho R}{6\varepsilon_{0}}

Electric field at point B = E_B = E_1A + E_2A
E_1A = Electric Field Due to solid sphere of radius R at point

B = \frac{\rho R}{3\varepsilon_{0}}

E_2A = Electric Field Due to solid sphere of radius R/2 (which having
charge density -

\rho

)

= - \frac{\rho R}{54\varepsilon_{0}}

Source: Grass-G / P-22 Physcis Paper 19 Dec Eng..docx

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