CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

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?


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 A kqa2 towards F

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


flag
Suggest Corrections
thumbs-up
2
mid-banner-image
mid-banner-image