Five point charges, each of value +q, are placed on five vertices of a regular hexagon of side L. The magnitude of the force on a point charge of value –q placed at the centre of the hexagon (in newton) is
A
Zero
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
√3q24πϵ0L2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
q24πϵ0L2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
q24√3πϵ0L2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cq24πϵ0L2 Suppose six charges were placed at the corners 1,2,3,4,5 and 6 of hexagon would have been zero.
Rightarrow⇁F01+⇁F02+⇁F03+⇁F04+⇁F05+⇁F06=0 ⇒⇁F01+⇁F02+⇁F03+⇁F04+⇁F05=−⇁F06 ⇒∣∣∣ResultantofforcesactingonchargeplacedatthecentreOofhexagon∣∣∣=|⇁F06| ⇒|Resultant|=q24πϵ0r2; since r=L{∵sin30∘=L2r} |Resultant|=q24πϵ0r2