Question

ABC is an equilateral triangle of side 10m and D is the mid-point of BC. Charges of +100,-100 and +75 micro-coulombs are placed at B,C,D respectively. Find the force on a +1 micro-coulomb charge placed at A.(Give your answer in S.I. unit).

Answer 1

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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 

$E_A = K  \frac{Q_A}{a^2} = 9 X 10^9X \frac{100 X 10^{-6}}{10^2}$

E(A) = 9000N/C

$E_B = K = \frac{Q_B}{a^2} = 9 X 10^9 X \frac{100 X 10^{-6}}{10^2}$

E(B) = 9000N/C

$E_D = K \frac{Q_D}{(AD)^2}$

$= 9 X 10^9 \frac{75 X 10^{-6}}{ \frac{3 X 10^2}{4} }$

$= 900 X 10 ^({9-6-2} )$

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 

$E_n = \sqrt{E_x^2 + E_y^2}$

$=\sqrt{(9000)^2 + (9000)^2}$

= 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
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Answer 2

Explanation:

both the pictures are the same answer hope it helps. the answer is continued in the second page photo after the angle is given