IIT JEE , AIEEE Forum - Physics , Electricity

determinedforiit

find the electric force on 2µC charge placed at the common centre O of two equilateral triangles each of side 10 cm,as shown in fig.(http://picasaweb.google.com/ankitmahapatra22/Desktop?authkey=LQhmO9JFSX8#)

electric charge on A(2µC),B(2µC),C(2µC),D(-2µC),E(-2µC)and F(-2µC).

ganesha1991

let the distance from the centre to vertex be r
r^2=100/3

now the force between A and 2 and between Band 2 adds up and is in the direction OD
similarly for B and E the force is towards OE and for C and F it is towards OF

now angle betwee angle DOE = angle EOF = 60
so fnet = F+Fcos60+Fcos60
= 2F
F=k2*2/ r^2
= 4k/ [100/3] = 3k/25

so fnet = 2*3k/25=6k/25

determinedforiit

the ans is 43.2N. now can anyone solve it???

determinedforiit

someone try it.

muskaanrelhan

Distance of the vertex from the centre of trianlgle= m

F=k q1 q2/r^2

Solving we get F= 10.8 N

Now net force on 2uC charge placed at the centre O= 2F + 2Fcos60 + 2Fcos60

[ forces by charges on A and E add up, forces by charges on F and C add up , forces by charges on A and D add up]

Net Force= 4*F

=4*10.8

=43.2 N

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