A) zero
B) \[\frac{8Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
C) \[\frac{6Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
D) \[\frac{4Q}{4\pi {{\varepsilon }_{0}}{{d}^{2}}}\]
Correct Answer: B
Solution :
Two equal and opposite charges are placed at a distance d. Electric field at centre due to + 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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