wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

5 equal charges q each are placed at the corners of a regular hexagon ABCDEF of side a as shown. Find the net electric field at the centre O of the hexagon?


A
5kqa2 towards F
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
kqa2 towards F
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
32kqa2 towards F
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
52kqa2 towards C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B kqa2 towards F
Inorder to solve, find electric field due to each charge at O and then add using principle of superposition.


Distance between O and B is a as side length of regular hexagon is given to be a.


Now EOB=kqa2 , directed radially 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.

Hence, net field will be only due to point charge at C.

EOC=kqa2 direction away from C i.e. towards F


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon