Question

Three point charges are arranged along the x axis. Charge q1 = -4.00 nC is located at x= 0.200 m and charge q2 = 2.00 nC is at x= -0.340 m . A positive point charge q3 is located at the origin.

What must the value of q3 be for the net force on this point charge to have magnitude 4.00 μN ?

Where along the x axis can q3 be placed and the net force on it be zero, other than the trivial answers of x=+∞ and x=−∞?

Answer 1

Three point charges are arranged along the x axis. figure can be shown below :

force due to charge q₁ on q₃ which is given as,

F₁ = k q₃ q₁ / r₁²    { eq. 1 }

where, k = 9 x 10⁹ Nm²C²

q₁ = charge on x-axis = - 4nC = - 4 x 10^-9 C

r₁ = distance between charge q₃ & q₁ = 0.2 m

insertiong the values in above eq.

F₁ = (9 x 10⁹ Nm²C²) q₃ ( x 10^-9 C) / (0.2 m)²

F₁ = 900 q₃                              { eq.2 }

force due to charge q₂ on q₃ which is given as,

F₂ = k q₃ q₂ / r₂² { eq. 3 }

where, q₂ = charge on x-axis = 2 nC = 2 x 10^-9 C

r₂ = distance between charge q₃ & q₂ = -0.34 m

inserting all these values in eq.3,

F₂ = (9 x 10⁹ Nm²C²) q₃ (2 x 10^-9 C) / (-0.34 m)²

F₂ = 156 q₃                                             { eq.4 }

Value of q₃ must be for the net force on this point charge to have magnitude 4 _$\mu$ N. The net force on the charge q₃ which is given as ::

F_net = F₁ + F₂

where, F_net = 4 x 10^-6 N

(4 x 10^-6 N) = (900 q₃) + (156 q₃)

(4 x 10^-6 N) = (1056 N) q₃

q₃ = (4 x 10^-6 N) / (1056 N)

q₃ = 0.00378 x 10^-6 C

q₃ = 3.78 x 10^-9 C

(b) the net force on it be zero. figure can be shown below :

F₁ = F₂

k q₃ q₂ / (r - r₂)² = k q₃ q₁ / (r + r₁)²

inserting the values in above eq.

(3.78 x 10^-9 C) (2 x 10^-9 C) / (r - 0.34m)² = (3.78 x 10^-9 C) (4 x 10^-9 C) / (r + 0.2m)²

2 / (r - 0.34m)² = 4 / (r + 0.2m)²

(r + 0.2m)² / (r - 0.34m)² = (4 / 2)

(r + 0.2m)² / (r - 0.34m)² = 2

(r + 0.2m) / (r - 0.34m) = 1.414

(r + 0.2m) = (1.414) (r - 0.34m)

(r + 0.2) = 1.414 r - 0.48

(0.414 r) = 0.68

r = 1.64 m

It means that charge q must be placed at x = -1.64 m