5 equal charges ‘q’ are placed at corners of a regular hexagon ABCDEF of side ‘a’ as shown. Find the net electric field at the centre O of the hexagon?
Find E due to each charge then add by principle of superposition.
Distance between O and B is ‘a’ as side length of regular hexagon is given to be ‘a’
Now →EOB=kqa2 , directedradially away from B
→EOE=kqa2 same magnitude but direction exactly opposite to →EOB
→EOE and →EOB are equal and opposite vectors and hence will cancel each other out.
Similarly →EOA and →EOD are equal and opposite so they will also have a zero net effect.
Net field will be only due to point charge at C
→EOC=kqa2 direction away from C i.e., towards F