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 :

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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