Question

The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed to each corner, as shown. The 4.00 microC charge experiences a net force due to the charges qA and q B. This net force points vertically downward and has a magnitude of 405 N. Determine the magnitudes and algebraic signs of the charges qA and q B.

Answer 1

Using Coulomb's Law and the symmetry of the equilateral triangle setup, the charges qA and qB must have opposite signs and their individual forces on the 4.00 µC charge can be calculated to satisfy the total net force of 405 N.

Step-by-step explanation:

To determine the magnitudes and signs of the charges qA and qB, we need to use Coulomb's Law, which states that the electric force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. Given that the 4.00 µC charge experiences a net force of 405 N vertically downward, we can deduce that the other two charges must have opposite signs since they need to pull the 4.00 µC charge in the same vertical direction.

Let's denote the unknown charges as qA and qB. Because the triangle is equilateral, the net force is vertical, which implies that the horizontal components of the forces due to qA and qB cancel each other out. Therefore, we can analyze each charge's influence separately. Applying Coulomb's Law, we calculate the vertical forces exerted by qA and qB on the 4.00 µC charge and equate their sum to the net force of 405 N to solve for the unknown charges.

Answer 2

`-2.60097* 10^(-6)\ C`

Step-by-step explanation:

k = Coulomb constant =
`8.99* 10^(9)\ Nm^2/C^2`

r = Distance between charges = 2 cm

The electric force is given by

`F=(k4* 10^(-6)q_A)/(r^2)`

Also

`F=(k4* 10^(-6)q_B)/(r^2)`

The sum of the force is given by

`\sum F=(k4* 10^(-6)q_A)/(r^2)cos 30+(k4* 10^(-6)q_B)/(r^2)cos 30\\\Rightarrow \sum F=2(k4* 10^(-6)q_A)/(r^2)cos 30\\\Rightarrow q_a=(\sum Fr^2)/(2* k4* 10^(-6)cos30)\\\Rightarrow q_a=(405* 0.02^2)/(2* 8.99* 10^9* 4* 10^(-6)cos30)\\\Rightarrow q_a=2.60097* 10^(-6)\ C`

Thus
`q_a=q_b=-2.60097* 10^(-6)\ C`.

The negative sign is because the force points downward which is taken as negative