A) \[8.6\times {{10}^{9}}N{{C}^{-1}}\]
B) \[4.1\times {{10}^{6}}\,N{{C}^{-1}}\]
C) \[3.6\,\times {{10}^{6}}N{{C}^{-1}}\]
D) \[4.6\times {{10}^{5}}N{{C}^{-1}}\]
Correct Answer: C
Solution :
(c) \[3.6\times {{10}^{6}}N{{C}^{-1}}\] Here, \[q=\pm 20\mu C=\pm .\,20\times {{10}^{-6}}C\], \[2a=10\,mm\,=10\times 10{{\,}^{-3}}m\], \[r=OP=10\,cm=10\times {{10}^{-2}}m\] \[\left| \overrightarrow{p} \right|=q\times 2a=20\times {{10}^{-6}}\times 10\times {{10}^{-3}}\] \[=2\times {{10}^{-7}}\,cm\] The electric field, along BP, \[\overrightarrow{E}=\frac{2\left| \overrightarrow{p} \right|r}{4\pi {{\varepsilon }_{0}}{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}\] As \[a<<r\] , \[\overrightarrow{E}=\frac{2\left| \overrightarrow{p} \right|}{4\pi {{\varepsilon }_{0}}{{r}^{3}}}=\frac{2\times 2\times {{10}^{-7}}\times 9\times {{10}^{9}}}{{{\left( 10\times {{10}^{-2}} \right)}^{3}}}\] \[=3.6\times {{10}^{6}}\,N{{C}^{-1}}\]You need to login to perform this action.
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