You run a comb through your hair and it becomes positively charged. Does its nass slightly increase or slightly decrease? Why? Where does the mass go to/come from?

Given two particles with Q= 2.70-μC charges as shown in the figure below and a particle with charge q = 1.31 x 10-18 C at the origin. (Note: Assume a reference level of potential V = 0 at r = ∞o.) Q x = -0.800 m 0 x = 0.800 m x Ⓡ (a) What is the net force (in N) exerted by the two 2.70-μC charges on the charge q? (Enter the magnitude.) N (b) What is the electric field (in N/C) at the origin due to the two 2.70-μC particles? (Enter the magnitude.) N/C (c) What is the electrical potential (in kV) at the origin due to the two 2.70-μC particles? kv (d) What If? What would be the change in electric potential energy (in J) of the system if the charge q were moved a distance d = 0.400 m closer to either of the 2.70-μC particles? J

A particle (charge = 3 µC) is released from rest at a point x %3D = 11.4 cm. If a 61.1-µC charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 94.1 cm? Round your answer to 2 decimal places.

A particle (charge= 2.1x10^-6 C) is released from rest at a point x=10.1 cm. If a 42.3x10^-6 C charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 92.8 cm?

Using the symmetry of the arrangement, determine the direction of the force on +q in the figure below, given that: Case I. qa=qb= +6.5 μC and qc = qd = +6.5 μC. (a) In your notebook, draw the forces on q due to qar qb²9c² (b) Due to symmetry the direction of the net force is F O qc net ✔N O qd Hint: For each force draw the x and y components. Some will add and some will cancel. (c) Calculate the magnitude of the force on the charge q, given that the square is 10.0 cm on a side and q = 2.3 μC. and 9d- X