JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 19)
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 :
_9th_January_Morning_Slot_en_19_1.png)
$${{17} \over {54}}$$
$${{18} \over {54}}$$
$${{18} \over {34}}$$
$${{21} \over {34}}$$
Explanation
$$\left| {\overrightarrow {{E_A}} } \right|$$ = (Ecomplete)center - (Eremoved)surface
= 0 - $$\left| {{{k\rho {4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}} \over {{{\left( {{R \over 2}} \right)}^2}}}} \right|$$
= $${k\rho {4 \over 3}\pi \left( {{R \over 2}} \right)}$$
$$\left| {\overrightarrow {{E_B}} } \right|$$ = (Ecomplete)surface - (Eremoved)r = $${{3r} \over 2}$$
= $${{{k\rho {4 \over 3}\pi {R^3}} \over {{R^2}}}}$$ - $${{{k\rho {4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}} \over {{{\left( {{{3R} \over 2}} \right)}^2}}}}$$
= $${k\rho {4 \over 3}\pi R}$$ - $${k\rho {4 \over 3}\pi {R \over {18}}}$$
= $${k\rho {4 \over 3}\pi \left( {{{17R} \over {18}}} \right)}$$
$$ \therefore $$ $${{\left| {\overrightarrow {{E_A}} } \right|} \over {\left| {\overrightarrow {{E_B}} } \right|}}$$ = $${{18} \over {34}}$$
= 0 - $$\left| {{{k\rho {4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}} \over {{{\left( {{R \over 2}} \right)}^2}}}} \right|$$
= $${k\rho {4 \over 3}\pi \left( {{R \over 2}} \right)}$$
$$\left| {\overrightarrow {{E_B}} } \right|$$ = (Ecomplete)surface - (Eremoved)r = $${{3r} \over 2}$$
= $${{{k\rho {4 \over 3}\pi {R^3}} \over {{R^2}}}}$$ - $${{{k\rho {4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}} \over {{{\left( {{{3R} \over 2}} \right)}^2}}}}$$
= $${k\rho {4 \over 3}\pi R}$$ - $${k\rho {4 \over 3}\pi {R \over {18}}}$$
= $${k\rho {4 \over 3}\pi \left( {{{17R} \over {18}}} \right)}$$
$$ \therefore $$ $${{\left| {\overrightarrow {{E_A}} } \right|} \over {\left| {\overrightarrow {{E_B}} } \right|}}$$ = $${{18} \over {34}}$$
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