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JEE Advanced2012PhysicsElectrostaticsActual

Six point charges are kept at the vertices of a regular hexagon of side L and centre O , as shown in the figure. Given that K= 1 4 ₀ q L² , which of the following statement(s) is (are) correct?

Options

  1. AThe electric field at O is 6 ~K along O D
  2. BThe potential at O is zero
  3. CThe potential at all points on the line P R is same
  4. DThe potential at all points on the line S T is same

Correct answer

A. The electric field at O is 6 ~K along O D

Step-by-step solution

Here aligned & | E_ A | 2 = | E_ B |= | E_ C |= | E_ D | 2 = | E_ E |= | E_ F |=K & E_ O =E_ A +E_ D + (E_ F +E_ C ) 60^ + (E_ B +E_ C ) 60^ =& 2 ~K +2 ~K +( K + K ) 1 2 +( K + K ) 1 2 =6 ~K aligned Electric potential at OV_ O = 1 4 ₀ L [2 q+q+q-q-q-2 q]=0 Potential at all points on the line PR is same not on line ST. P R is perpendicular bisector (the equatorial line) for the electric dipoles A B, F E and B C . Therefore the electric potential will be zero at any point on P R .

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