Chapter 22 Textbook Problems

Answers to odd problems are found in the back of the textbook

ELECTRIC POTENTIAL

Section 1Electric Potential Energy and1, 2, 3

The Electric Potential

Section 2Using the Electric Potential4, 7, 8

Section 3Calculating the Electric Potential9, 13, 14, 16

Section 5Connecting Potential and Field18, 19, 20, 21, 22

43, 44, 45(what would be the electron’s velocity as it passes through the protons if it was released from rest and allowed to move freely), 55

P21.2. Prepare: The electric potential energy developed between two chargesandseparated by a distance r may be determined by

Solve: Knowing (from the graph) that the electric potential energy developed between the two charges when they are separated by a distance ofis and knowing the value of one charge (q1 we can determine the value of the second chargeby

Assess: This is a reasonable value for the second charge. Since one charge is positive and the other negative, they are bound together, which is consistent with the fact that the electric potential energy is negative. As a check on our work we could pull another set of values from the graph and repeat the calculation.

P21.4. Prepare: Mechanical energy is conserved. The potential energy is determined by the electric potential and is given by U  qV. The figure shows a before-and-after pictorial representation of an electron moving through a potential difference of 1000 V. A negative charge speeds up as it moves into a region of higher potential (U  K). We also note that 1 eV  1.60  10–19 J.

Solve: (a) Conservation of energy, expressed in terms of the electric potential V, is

(b)

(c) Becauseso

 1.9  107 m/s

Assess: The electric potential difference of 1000 V already existed in space due to other charges or sources. The electron of our problem has nothing to do with creating the potential.

P21.8. Prepare: Energy is conserved. The potential energy is determined by the electric potential. The figure shows a before-and-after pictorial representation of a proton moving through a potential difference.

Solve: (a) Because the proton is a positive charge and it slows down as it travels, it must be moving from a region
of lower potential to a region of higher potential.

(b) Using the conservation of energy equation,

(c)

Assess: A positive V confirms that the proton moves into a higher potential region.

P21.14. Prepare: Please refer to Figure P21.14. The charge is a point charge. We will use Equations 21.9 and 21.1 to calculate the potential and the potential energy of the charge.

Solve: (a) The electric potential of the point charge q is

For points A and B, r 0.01 m. Thus,

For point C, r 0.02 m and

(b) The potential energy of a chargeat a point where the electric potential is V is UThe expression for the potential in part (a) assumes that we have chosen V 0 V to be the potential at r. So, we are obtaining potential/potential energy relative to a zero of potential/potential energy at infinity. Thus,

(c) The potential differences are

VABVBVA 1800 V 1800 V  0 V VBCVCVB 900 V  1800 V 900 V

Assess: Clearly VAVB and VC and VA, so, as expected, and is negative.

P21.16. Prepare: Please refer to Figure P21.16. The net potential is the sum of the potentials due to each charge given by Equation 21.9.

Solve: The potential at the dot is

Assess: Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.

P21.20. Prepare: We will use Equation 21.16 to find the electric field.

Solve: The electric potential difference V between two points in a uniform electric field is:

V Exd

1000 V  (1000 V) Ex[1.0 m (1.0 m)]Ex 1000 V/m

Assess: Electric field points from higher potential to lower potential.

P21.22. Prepare: Please refer to Figure P21.22. Three equipotential surfaces at potentials of 200 V, 0 V, and 200 V are shown. The electric field is perpendicular to the equipotential lines and points “downhill.”

Solve: The electric field perpendicular to the equipotential surfaces is

The electric field vector is in the third quadrant, 45 below the negative x-axis. That is,

Assess: As obtained above, electric field points from higher potential to lower potential.

P21.44. Prepare: Please refer to Figure P21.44. For any path that starts and ends at the same point,

Solve: For the loop shown in the figure:

V12V23V34V41 30 V  50 V V34 – 60 V  0 VV3420 V

Or, V43 +20 V. That is, the potential at point 4 is less than the potential at point 3.