Question

Three point charges +2*10^-6 c , -3*10^-6 and -3*10^-6 are kept at the vertices A , B and C respectively of an equilateral triangle of side 20 cm . What should be the sign and magnitude of the charge to be placed at the mid point of side BC so that the charge at A remains in equilibrium?

Answer 1

I don't no answer .......

Answer 2

here in the picture
F1 = force on the A by the B
F2 = force on the A by the C
F3 = resultant force of B & C
F4 = force on the A by the D


also it is give that
charge on A = 2*10^-6 c
charge on B = -3*10^-6c
charge on C = -3*10^-6c
angle between each force is 60degree

let the charge in D is q

now we will calculate F1 & F2

F1 = 1/4πEó q1q2/ r^2

F1 = [(9*10^9)*(2*10^-6)*(3*10^-6)]/(0.2)^2

F1 = 135N

now we will calculate F2
since the value of charge and distance for F2 is same as F1 so
F1 = F2

so,
F2 = 135N

now the resultant of F1 and F2 is
by the vector addition

F3 = √[(F1)^2 + (F2)^2 + 2(F1)(F2)cosø]

(since F1 = F2)

F3 = √[(F1)^2 + (F1)^2 + 2(F1)(F1)cos(60)]

F3 = √[ 2(F1)^2 + 2(F1)^2 *1/2]

F3 = √[3(F1)^2]

F3 = F1√(3)

F3 = 135√(3)N

since the resultant force is towards the D so we need a same(positive) charge to repel it and hence charge A will remain at equilibrum. Because like charges repel each other

so.
F4 = F3

1/4πEò * 2*10^-6*q/r^2 = F3

(9*10^9)*(2*10^-6)*(q) / [√(3)/2 * 20]^2 = 135√(3)N

(9*10^9)*(2*10^-6)*(q) / [√(3) * 10]^2 = 135√(3)N

(9*2*10^3)*(q) / 3*100 = 135√(3)N

60 *q = 135√(3)N

q = 135√(3) / 60

q = 3.897 c

so this is the magnitude and the sign will be positive.

please check this if u find any mistake please let me know.

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