Electric Fields Practice Test Circle the best answer. 1. If an object has zero net charge, what does this mean? a) there are extra protons vs. electrons b) there are extra electrons vs. protons c) the number of neutrons equals the number of electrons d) the number of protons equals the number of neutrons e) none of the above are correct 2. Insulators cannot conduct electricity, yet a balloon can be made to stick to a wall. This is possible because a) the neutrons in the wall polarize because the nearby balloon is positively charged b) the atoms in the wall polarize because the nearby balloon is negatively charged c) the electrons in the wall are attracted to the neutrons in the balloon d) the neutrons in the balloon polarize due to the fundamental rule e) none of the above are correct 3. According to the laws of electrostatics, protons do not want to be near each other and will fly apart. Yet in the nucleus protons are stuck right beside each other. What keeps them together? a) the electrons acting like a nuclear “super-glue” b) the strong nuclear force c) the weak nuclear force d) electromagnetism e) none of the above are correct 4. If you move a charged particle against an electric field, like we did in a capacitor, what value have you increased for this particle? a) voltage b) charge c) mass d) gravitational potential energy e) electric field

5. An object has a charge of 2.5 x 10-5 μC. What is the electric field strength 0.5 cm away from this object?

6. What is the electrostatic force between two objects that are 5.7 x 10-4 m apart with one object having a charge of 9.5 mC and the other a charge of 7.4 mC?

7. A positive charge of 5.0 μC is resting against the negative plate of a capacitor. The electric field strength between the two plates is 2.0 x 105 N/C. The distance between the plates is 10.0 cm. a) How much force is required to begin moving the particle to the positive plate? b) How much work is required to move the particle to the positive plate? c) How much voltage is given to this particle?

8. You have two electrons. a) What is the electrostatic force between these two objects if they are 2 cm apart? b) If the electrons were moved to a distance of 5 cm apart, what would the force now be?

9. What is the electric field intensity in a situation where a negative charge of -2.4 x 10-6 μC experiences a force of 5 x 10-2 N?

10. If you have 1.0 x 103 C, how many electrons would this be equal to?

11. If two points are 0.10 m apart and there is 80.0 V between them, what is the electric field strength?

12. The magnesium atom contains twelve protons in the nucleus. When magnesium is ionized to the point where only two electrons remain, the remaining electron pair orbits the nucleus at a radius of 3.45 x 10-10 m. a) What is the magnitude of the electric field created at the location of the electron pair by the positively charged nucleus? b) What is the force acting on the electron pair?

13. What is the total charge of all the electrons in 1.92 kg of copper sulfate (CuSO4)?

14.

There are three positively charged points, each with a charge of 5.57 x 10-2 C, fixed at the vertices of an equilateral inverted triangle whose sides are 75 cm long. Find the net force on the southernmost point.

15. Using the diagram below, find a) the magnitude and direction of the force on q1 b) the magnitude of the electric field at q1.

16. Four point charges A, B, C, and D, each with a magnitude of 2.50 x 10-1 C are found at the corners of a square whose sides are 4.92 m long. The polarity and position of the charges are shown in the diagram below. Determine the magnitude of the force acting on charge A.

Bonus (+4) The diagram below shows two balls moving towards each other and about to collide elastically. Using the magnitudes of velocity and the angles below, determine the magnitude and direction of the momentum of m2 after the collision.

Formula Sheet F = kq1q2/d

2

E = kQ/d2 E = F/q W = Fd V = W/q V = Ed Avogadro's Number = 6.02 x 1023 q of e- = -1.602 x 10-19 C q of p+ = 1.602 x 10-19 C k = 9.0 x 109 1 C = charge on 6.24 x 1018 electrons # mol = mass/molar mass Bonus Formulae p = mv pxi = p1xi + p2xi pxf = p1xf + p2xf pxi = pxf pyi = p1yi + p2yi pyf = p1yf + p2yf pyi = pyf

Solutions EBBA 5. 9 x 103 6. 1.9 x 1012 7. a) 1.0 b) 0.10 c) 2.0 x 104 8. a) 6 x 10-25 b) 9 x 10-26 9. -2 x 1010 10.6.2 x 1021 11.8.0 x 102 12.a) 1.45 x 1011 b) -4.66 x 10-8 13.-8.93 x 107 14.-8.6 x 107 15.a) 5.0 at 42o N of E b) 8.9 x 105 16.3.49 x 107 Bonus: 9.55 x 10-1 kgm/s at 37.8o N of E