Question

Three point charges, two positive and one negative, each having a magnitude of 26 micro-C are placed at the vertices of an equilateral triangle (48 cm on a side). What is the magnitude of the electrostatic force on the negative charge?

Answer 1

84.44N

Step-by-step explanation:

Hi!

The force F between two charges q₁ and q₂ at a distance r from each other is given by Coulomb's law:

`F = k_c (q_1 q_2)/(r^2)`

The force on the negative charge q₁ is the sum of the forces from the other two charges. This forces have equal magnitude as both distances are 48cm. The magnitud is:

`F_(1,2) =F_(1,3) = -k_c((26\mu C)^2)/((48cm)^2)=-9*10^9Nm^2C^(-2)*0.54*(10^(-12)C^2)/(10^(-4)m^2)=-48.75N\\`

(negative means attractive)

The sum of the forces, because of symmetry reasons actos along line L (see the figure), and its magnitud is:

`F = 2*48.75*\cos(30\º)N = 84.44N`