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Question

 A solid conducting sphere of radius $$2.00 cm$$ has a charge of $$8.00 \mu C$$. A conducting spherical shell of inner radius $$4.00 cm$$ and outer radius $$5.00 cm$$ is concentric with the solid sphere and has a charge of $$-4.00 \mu C$$. Find the electric field at (a) $$r =1.00 cm$$,(b) $$r=3.00 cm$$, (c) $$r =4.50 cm$$, and (d) $$r= 7.00 cm$$ from the center of this charge configuration.


Solution

(a) $$\overset{\rightarrow}{E}=0$$
(b) $$E=\dfrac{k_{e}Q}{r^{2}}=\dfrac{(8.99\times 10^{9}N\cdot m^{2}/C^{2})(8.00\times 10^{-6}C)}{(0.0300m)^{2}}=7.99\times 10^{7}N/C$$
$$\overset{\rightarrow}{E}=79.9MN/C$$ radially outward 
(c) $$\overset{\rightarrow}{E}=0$$
(d) $$E=\dfrac{k_{e}Q}{r^{2}}=\dfrac{(8.99\times 10^{9}N\cdot m^{2}/C^{2})(4.00\times 10^{-6}C)}{(0.0700m)^{2}}=7.34\times 10^{6}N/C$$
$$\overset{\rightarrow}{E}=7.34MN/C$$ radially outward

Physics
NCERT
Standard XII

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