question_answer

The electric intensity due to a dipole of length 10 cm and having a charge of \[500\,\mu C\] at a point on the axis at a distance 20 cm from one of the charges in air is :

A)  \[20.5\times {{10}^{7}}N/C\]

B)  \[18.28\,\,N/C\]

C)  \[6.25\times {{10}^{7}}N/C\]

D)  none of these

Correct Answer: C

Solution :

Key Idea: Dipole is made up of a system of two equal and opposite charges separated by a distance \[2\,l\]. Let \[{{E}_{1}}\] and \[{{E}_{2}}\]be the intensities due to charges + q and - q of dipole. \[\therefore \] \[{{E}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{(r-l)}^{2}}}\] \[{{E}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{(r-l)}^{2}}}\] Resultant, \[\overrightarrow{E}={{\overrightarrow{E}}_{1}}-{{\overrightarrow{E}}_{2}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{2\,2\,ql)r}{{{({{r}^{2}}-{{l}^{2}})}^{2}}} \right]\] where \[2\,ql\] is electric dipole moment. Given, \[2\,l=10\,cm=0.1\,m\] \[\therefore \] \[l=0.05\,m\] \[q=500\,\mu C=500\times {{10}^{6}}C,\,r=20+5=25\,cm\] \[=0.25\,m\] Note: The direction of the resultant field, is along the axis of the dipole from the negative charge towards positive charge.