Three point charges are arranged as shown in the figure below. Find the magnitude and direction of the electric force on the particle q = 4.80 nC at the origin. (Let r12 = 0.250 m.)

The figure below shows two point charges, 𝑄1 = +7.50 nC and 𝑄2 = −8.30 nC, separated bya distance of 5.00 mm.Calculate(a) the force acted upon 𝑄1.(b) the electric field strength at the centre between the point charges.(c) the electric potential at the centre between the point charges.

Consider three charges at points 𝑥1, 𝑦1 = (2.6 m, 1.2 m), 𝑥2, 𝑦2 = (3.7 m, 7.8 m), 𝑥3, 𝑦3 = (9.3 m, 5.2 m) with charges 𝑞1 = 7 nC, 𝑞2 = −9.6 nC, 𝑞3 = 4.7 nC. What is the magnitude of the electric field at the origin (𝑥 = 0, 𝑦 = 0) in N/C? (10 marks)

A 17.3-nC point charge is placed on the x-axis at x = 4.05 m, and a −35.5-nC point charge is placed on the y-axis at y = −3.57 m. What is the direction of the electric field at the origin? (Give your answer in degrees relative to the positive x-axis.)

A 4.40 μC charge is subject to a 6.00 mN force due to an electric field. What is the magnitude of the electric field at the location of the charge?  *1 point7.31 kN/C26.4 kN/C1.364 kN/C4.364 kN/C

A particle with positive charge q = 2.08 10-18 C moves with a velocity v = (4î + 4ĵ − ) m/s through a region where both a uniform magnetic field and a uniform electric field exist.(a)Calculate the total force on the moving particle, taking B = (4î + 3ĵ + ) T and E = (3î − ĵ − 2) V/m. (Give your answers in N for each component.)Fx = NFy = NFz = N(b)What angle does the force vector make with the positive x-axis? (Give your answer in degrees counterclockwise from the +x-axis.) ° counterclockwise from the +x-axis(c)What If? For what vector electric field would the total force on the particle be zero? (Give your answers in V/m for each component.)Ex = V/mEy = V/mEz = V/m