A) zero
B) \[\frac{8\,Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
C) \[\frac{6\,Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
D) \[\frac{4\,Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
Correct Answer: B
Solution :
Two equal and opposite charges are placed at a distance d. Electric field at centre due to \[+\text{ }Q\]charge \[({{E}_{1}})=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{(Q)}{{{(d/2)}^{2}}}\] Similarly, electric field due to \[-Q\] charge \[({{E}_{2}})=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{(Q)}{{{(d/2)}^{2}}}\] Therefore, net electric field at point \[E={{E}_{1}}+{{E}_{2}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{4Q}{{{d}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{4Q}{{{d}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{8Q}{{{d}^{2}}}\]You need to login to perform this action.
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