question_answer

In the following four situations charged particles are at equal distance from the origin. Arrange them the magnitude of the net electric field at origin greatest first  

(i)	(ii)
(iii)	(iv)

 

A) \[\left( i \right)>\left( ii \right)>\left( m \right)>\left( iv \right)\]

B) \[\left( ii \right)>\left( i \right)>\left( iii \right)>\left( iv \right)\]

C) \[\left( i \right)>\left( iii \right)>\left( ii \right)>\left( iv \right)\]

D) \[\left( iv \right)>\left( iii \right)>\left( ii \right)>\left( i \right)\]

Correct Answer: C

Solution :

If electric field due to charge \[\left| q \right|\] at origin is E. electric field due to charges \[\left| 2q \right|,\,\left| 3q \right|,\,\left| 4q \right|\,and\,\left| 5q \right|\] are respectively 2E, 3E, 4E and 5E (i) (ii) (iii)   (iv) 0 \[{{E}_{(i)}}=\,\,\sqrt{{{\left( 5E \right)}^{2}}+{{\left( 5E \right)}^{2}}}\,\,=\,\,5\sqrt{2}\,E\] \[{{E}_{(ii)}}=\,\,\sqrt{{{\left( 3E \right)}^{2}}+{{\left( 3E \right)}^{2}}}\,\,=\,\,3\sqrt{2}\,E\] \[{{E}_{(iii)}}=\,\,4E+2E=6E\,\,and\,\,{{E}_{(iv)}}\,=3E+E=4E\] \[\Rightarrow \,\,\,\,\,\,\,\,\,{{E}_{(i)}}>{{E}_{(iii)}}>{{E}_{(ii)}}>{{E}_{(iv)}}\]