Physics 2: Electric Forces and Fields
Two point charges, q = 6Q and q = -2Q, are separated by a distance, d. The attractive force between these two particles is 20 N. If the distanced is halved, so d = d/2, what is the new force between the particles?
In the figure below, A = 3 cm, B = 1 cm, q = +15 nC, q = -5 nC, and q is -10 nC. Find the magnitude and direction of the net electrostatic force on q
A proton ( 1.6 x 10 C ) is placed 2.19 x 10 m from point A. Find the electric field at point A. Imagine a proton is put at point A. Find the force that acts on the proton.
Two parallel charged plates, one with a surface charge density of = 40 and another, = -60 . Find the total electric field between the plates. (note this is not a capacitor)
The electron gun in a television tube is used to accelerate electrons from rest to 3.0 x 10 m/s within a distance of 2.0 cm. What electric field is required? Should the electric field be in the same or opposite direction of the electrons motion?
The figure below shows a single dipole of charges 5 nC, with a distance s = 0.002 m between the poles. Point A is a distance, d = 0.1 m, and is perpendicular to the dipole. Point B is a vertical distance, d = 0.1 m, and is parallel to the dipole. Find the electric field at points A and B.
Two dipoles are placed a distance, d = 0.05 m, from the point A (labeled in red). The distance between dipoles is s = 0.01 m. Each dipole consists of a positive and negative charge of 2 nC. Find the electric field at point A.
Two concentric loops are shown below, with a distance between them Y = 0.4 m, a radius Z = 0.1 m. The loop on the left has a charge of +9 nC and the loop on the right has a charge of -9 nC. Find the electric field at point X, which is directly in the middle of the loops.
Given the 3 dimensional electric field, = 200 + 300 + 400 , and an area vector = 0.2 m + 0.3 m , find the electric flux. (this is a calculus based physics problem)