Problem 1: Common static electricity involves charges in the range between nanocoulombs and microcoulombs. Part (a) How many electrons are needed to form a charge of Q1 = =1 nC? N= Part (b) How many electrons must be removed from a neutral object to leave a net charge of Q2 = 1.1 UC? N= Problem 4: An archaeon has 6.5 x 1016 protons and a net charge of 9.7pC. Part (a) How many fewer electrons are there than protons? No - Ne = Part (b) If you paired them up, what fraction of the protons would have no electrons? (No - Ne) / No = Problem 6: Common transparent adhesive tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges as an approximation, calculate the magnitude of each charge, in coulombs, if the electrostatic force is great enough to support the weight of a 0.75 mg piece of tape held 0.95 cm above another. q = Problem 7: Two balloons have equal and opposite charges. Balloon one has /= 105 excess electrons. The balloons are separated by d= 2.9 m and each electron has a negative charge of e = 1.602 x 10-19 C. Part (a) What is the charge on balloon two, Q2, in C? Q2 = Part (b) What is the magnitude of the force (in N) of balloon one on balloon two using the variables provided and the Coulomb constant k (k = 8.988 x 109 N m /C2). F12 = Problem 8: A positively charged particle Q101 = 25 nC is held fixed at the origin. A negatively charged particle Q292 = -16 nC of mass me = 4.5 ug is located a distance de = 45 cm Q d from the positively charged particle along the positive X-axis. Otheexpertta.com Part (a) What is the magnitude of the electric force in newtons that acts on the charge Q2? F12 = Part (c) What is the magnitude of the acceleration, a in m/s', experienced by charge Q2? a = Problem 9: Suppose you had a point charge of 4.6 mC. Calculate the magnitude of the electric field, in newtons per coulomb, at a point 2.2m away from this charge. Problem 10: Suppose a point charge creates a 12500 N/C electric field at a distance of 0.65 m. Part (a) What is the magnitude of the point charge in coulombs? | Q1 = Part (b) What is the strength of the field, in newtons per coulomb, at a distance of 10 m? E=Problem 11: The figure shows the electric field lines near two charges q1 (red, on the left) and q2 (blue, on the right). Otheexpertta.com Part (a) What is the ratio of their charges? q/ q2 = Problem 13: Point charges located at 3, 8, and 11 cm along the x-axis are represented in the figure. 10 x (cm) +q -2q Part (a) Find the electric field at x1 = 5.00 cm in N/C, given that q = 1.00 UC . E = Part (b) At what position between 3.00 and 8.00 cm is the total electric field the same as that for -2q alone in cm? X = theexpertta.com Part (c) At what position to the right of 11.0 cm is the total electric field zero in cm, other than at infinity? X3 = Problem 14: A simple and common technique for accelerating electrons is shown in the figure, which depicts a uniform electric field between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to pass through. Randomized Variables Otheexpertta.com E= 2.35 x 104 N/C Calculate the horizontal component of the electron's acceleration if the field strength is 2.35 x 104 N/C. Express your answer in meters per second squared, and assume the electric field is pointing in the negative x-direction as shown in the figure. ax = Problem 15: You are told that in a thundercloud, a 0.75 g raindrop can acquire a charge of 1.1 mC. Part (a) Assume that two such raindrops are separated by 2.9 cm. What is the acceleration, in meters per square second, of each raindrop away from each other