Question 431: Electrostatics & Capacitance

Question

Consider a sphere of radius R which carries a
uniform charge density $\rho $. If a sphere of radius ${{R \over 2}}$ is carved out of it, as shown, the ratio ${{\left| {\overrightarrow {{E_A}} } \right|} \over {\left| {\overrightarrow {{E_B}} } \right|}}$ of magnitude of electric field ${\overrightarrow {{E_A}} }$ and ${\overrightarrow {{E_B}} }$,
respectively, at points A and B due to the
remaining portion is :
![](https://imagex.cdn.examgoal.net/7k9k2k5i7j5a8/b0881c56-655c-499f-9019-1fbb94bb2a23/7942ef10-392e-11ea-92dd-4b9c3519996f/file-7k9k2k5i7j5a9-206w.jpg)

Accepted Answer

$\left| {\overrightarrow {{E_A}} } ight|$ = (E
complete
)
center
- (E
removed
)
surface
= 0 - $\left| {{{kho {4 \over 3}\pi {{\left( {{R \over 2}} ight)}^3}} \over {{{\left( {{R \over 2}} ight)}^2}}}} ight|$
= ${kho {4 \over 3}\pi \left( {{R \over 2}} ight)}$
$\left| {\overrightarrow {{E_B}} } ight|$ = (E
complete
)
surface
- (E
removed
)
r = ${{3r} \over 2}$
= ${{{kho {4 \over 3}\pi {R^3}} \over {{R^2}}}}$ - ${{{kho {4 \over 3}\pi {{\left( {{R \over 2}} ight)}^3}} \over {{{\left( {{{3R} \over 2}} ight)}^2}}}}$
= ${kho {4 \over 3}\pi R}$ - ${kho {4 \over 3}\pi {R \over {18}}}$
= ${kho {4 \over 3}\pi \left( {{{17R} \over {18}}} ight)}$
$ 	herefore $ ${{\left| {\overrightarrow {{E_A}} } ight|} \over {\left| {\overrightarrow {{E_B}} } ight|}}$ = ${{18} \over {34}}$

Answer

${{17} \over {54}}$

Answer

${{18} \over {54}}$

Answer

${{18} \over {34}}$

Answer

${{21} \over {34}}$

Electrostatics & Capacitance

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