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

JEE MainPhysicsToughElectrostatics

Consider a sphere of radius R which carries a uniform charge
density $\rho$ . If a sphere of radius $\frac{R}{2}$ is carved out of it,
as shown, the ratio
$|\frac{{\overset{\rightarrow}{E}}_{A}}{{\overset{\rightarrow}{E}}_{B}}|$
of magnitude of electric field
${\overset{\rightarrow}{E}}_{A}$ and ${\overset{\rightarrow}{E}}_{B}$ ,
respectively, at points 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(3\varepsilon_{0})}

|E_{A}| = \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}}

E_{B} = E_{1A} + E_{2A} = \frac{\rho R}{3\varepsilon_{0}} - \frac{\rho R}{54\varepsilon_{0}} = \frac{17\rho R}{54\varepsilon_{0}}

\frac{|E_{A}|}{|E_{B}|} = \frac{9}{17}

Source: Grass-G / P-19 Physics paper 28 Nov Fourth.docx

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