Question

ABC is an equilateral triangle of side 10 m and D is the midpoint of BC. Charges of 100, - 100 and 75 microcoulomb are placed at B, C and D respectively. Find the force on a 1 microcoulomb charge placed at A.

Answer 1

____________________________________________________________At The Point A 
Net field along x - axis 
E(x) = E(B) cos 60  + E(c) cos 60 
E(y) = E(D) + E(B) sin 60  - E(c) sin 60
Given that , a = 10 m 
Q(A) = + 1 μC
Q(B) = 100 μC
Q(c)  =  -100 μC
Q(D) = 75 μC 
k = 9 x 10⁹ N M ²C⁻²
Electric field ( magnitude)  are 


E(A) = 9000N/C


E(B) = 9000N/C






E(D) = 9000N/C
So , E(x)  = 9000 cos  60  + 9000 cos 60  
= [9000 x (1/2) ]x 2 = 9000N/C

E(y) =  9000 + 9000 sin 60 -  9000 sin 60 
= 9000N/C
So , net field of point A 




= 9000√2
E(n) = 12727.9N/C
So , Net force acting on point charger of (+1μC) is Given By
F = q E(n)      =(1 x 10⁻⁶) 12727.9 
F = 12.73 x 10⁻³N____________________________________________________________