Question

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge q1 is at x = 2 cm. What is q1 (magnitude and sign) if the net force on q3 is zero.

Answer 1

q₁ = + 1.25 nC

Step-by-step explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

`(k*q_(1)*q_3 )/((d_(13))^(2)  ) = (k*q_(2)*q_3 )/((d_(23))^(2)  )`

We divide by (k * q3) on both sides of the equation

`(q_(1) )/((d_(13))^(2) ) = (q_(2) )/((d_(23))^(2) )`

`q_(1) = (q_(2)*(d_(13))^(2)   )/((d_(23) )^(2)  )`

`q_(1) = (5*(2)^(2) )/((4 )^(2)  )`

q₁ = + 1.25 nC