A) 7.2 N towards the line charge
B) 6.6 N away from the line charge
C) 0.6 N away from the line charge
D) 0.6 N towards the line charge.
Correct Answer: D
Solution :
[d]: The electric field at a distance r from the line charge of linear density \[\lambda \]is given by\[E=\frac{\lambda }{2\pi {{\varepsilon }_{0}}r}\] Hence, the field at the negative charge, \[{{E}_{1}}=\frac{(4.0\times {{10}^{-4}})(2\times 9\times {{10}^{9}})}{0.02}=3.6\times {{10}^{8}}N{{C}^{-1}}\] The force on the negative charge, \[{{E}_{1}}=(3.6\times {{10}^{8}})(2.0\times {{10}^{-8}})\] = 7.2 N towards the line charge Similarly, the field at the positive charge, i.e., at r = 0.022 m is\[{{E}_{2}}=3.3\times {{10}^{8}}N{{C}^{-1}}\]. The force on the positive charge, \[{{F}_{2}}=(3.3\times {{10}^{8}})\times (2.0\times {{10}^{-}}^{8})\] = 6.6 N away from the line charge. Hence, the net force on the dipole = 7.2 N - 6.6 N = 0.6 N towards the line charge
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