First slide

Electric field

Question

Consider a sphere of radius R which carries a uniform charge density $ρ$ . If a sphere of radius $\frac{R}{2}$ is carved out of it, as shown , the ratio $\frac{|{\vec{E}}_{A}|}{|{\vec{E}}_{B}|}$ of magnitude of electric field ${\vec{E}}_{A}$ and ${\vec{E}}_{B}$ . respectively, at points A and B due to the remaining portion is:

Difficult

A

$\frac{18}{34}$
B

$\frac{18}{54}$
C

$\frac{17}{54}$
D

$\frac{21}{34}$

Solution

$\begin{array}{l} A s s u m e c h a r g e d e n s i t y i n s i d e c a v i t y a n d o u t s i d e c a v i t y i s + ρ a n d - ρ r e s p e c t i v e l y . \\ E_{i n} = \frac{ρ r}{3 \in_{0}} (r < R) \\ E_{o u t} = (\frac{ρ R^{3}}{3 \in_{0}}) \frac{1}{r^{2}} (r > R) \\ P o i n t A i s i n s i d e b o t h s p h e r e s, \\ E_{A} = E_{1} + E_{2} = \frac{ρ}{3 ε_{0}} (0) + \frac{- ρ}{3 ε_{0}} (\frac{R}{2}) = \frac{- ρ R}{6 ε_{0}} \\ P o i n t B i n s i d e f o r (1) a n d o u t s i d e f o r (2) \\ E_{B} = E_{1} + E_{2} = \frac{ρ R}{3 ε_{0}} + \frac{- ρ}{3 ε_{0}} \frac{{(\frac{R}{2})}^{3}}{{(\frac{3 R}{2})}^{2}} = \frac{17 ρ R}{54 ε_{0}} \\ R a t i o = \frac{E_{A}}{E_{B}} = \frac{18}{34} \end{array}$

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