question_answer

Two charges \[+20\mu C\]  and \[-20\,\mu \,C\] are placed 10 mm apart. The electric field at point P, on the axis of the dipole 10 cm away from its centre O on the side of the positive charge is:

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 \[a<<r\] , \[\overrightarrow{E}=\frac{2\left| \overrightarrow{p} \right|}{4\pi {{\varepsilon }_{0}}{{r}^{3}}}=\frac{2\times 2\times {{10}^{-7}}\times 9\times {{10}^{9}}}{{{\left( 10\times {{10}^{-2}} \right)}^{3}}}\] \[=3.6\times {{10}^{6}}\,N{{C}^{-1}}\]