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Chapter 23: Electric Potential . Section 23-1: Potential Difference. The voltage between the cathode and the screen of a television set is 22 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed just before it hits the screen?. 8.8 × 10 7 m/s
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Chapter 23: Electric Potential Section 23-1: Potential Difference
The voltage between the cathode and the screen of a television set is 22 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed just before it hits the screen? • 8.8 × 107 m/s • 2.8 × 106 m/s • 6.2 × 107 m/s • 7.7 × 1015 m/s • 5.3 × 107 m/s
The voltage between the cathode and the screen of a television set is 22 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed just before it hits the screen? • 8.8 × 107 m/s • 2.8 × 106 m/s • 6.2 × 107 m/s • 7.7 × 1015 m/s • 5.3 × 107 m/s
The electric field in a region is given by E = 2x2i + 3yj where the units are in V/m. What is the change in electric potential from the origin to (x, y) = (2, 0) m? • 8 V • –8 V • –16/3 V • –24/3 V • 11 V
The electric field in a region is given by E = 2x2i + 3yj where the units are in V/m. What is the change in electric potential from the origin to (x, y) = (2, 0) m? • 8 V • –8 V • –16/3 V • –24/3 V • 11 V
A lithium nucleus with a charge of +3e and a mass of 7 u, and an alpha particle with a charge of +2e and a mass of 4 u, are at rest. They could be accelerated to the same kinetic energy by • accelerating them through the same electrical potential difference. • accelerating the alpha particle through V volts and the lithium nucleus through 2V/3 volts. • accelerating the alpha particle through V volts and the lithium nucleus through 7V/4 volts. • accelerating the alpha particle through V volts and the lithium nucleus through 7V/6 volts. • none of these procedures.
A lithium nucleus with a charge of +3e and a mass of 7 u, and an alpha particle with a charge of +2e and a mass of 4 u, are at rest. They could be accelerated to the same kinetic energy by • accelerating them through the same electrical potential difference. • accelerating the alpha particle through V volts and the lithium nucleus through 2V/3 volts. • accelerating the alpha particle through V volts and the lithium nucleus through 7V/4 volts. • accelerating the alpha particle through V volts and the lithium nucleus through 7V/6 volts. • none of these procedures.
The concept of difference in electric potential is most closely associated with • the mechanical force on an electron. • the number of atoms in one gram-atom. • the charge on one electron. • the resistance of a certain specified column of mercury. • the work per unit quantity of electric charge.
The concept of difference in electric potential is most closely associated with • the mechanical force on an electron. • the number of atoms in one gram-atom. • the charge on one electron. • the resistance of a certain specified column of mercury. • the work per unit quantity of electric charge.
Charges Q and q (Q≠q), separated by a distance d, produce a potential VP = 0 at point P. This means that • no force is acting on a test charge placed at point P. • Q and q must have the same sign. • the electric field must be zero at point P. • the net work in bringing Q to distance d from q is zero. • the net work needed to bring a charge from infinity to point P is zero.
Charges Q and q (Q≠q), separated by a distance d, produce a potential VP = 0 at point P. This means that • no force is acting on a test charge placed at point P. • Q and q must have the same sign. • the electric field must be zero at point P. • the net work in bringing Q to distance d from q is zero. • the net work needed to bring a charge from infinity to point P is zero.
When +2.0 C of charge moves at constant speed from a point with zero potential to a point with potential +6.0 V, the amount of work done is • 2 J. • 3 J. • 6 J. • 12 J. • 24 J.
When +2.0 C of charge moves at constant speed from a point with zero potential to a point with potential +6.0 V, the amount of work done is • 2 J. • 3 J. • 6 J. • 12 J. • 24 J.
The electron volt is a unit of • capacitance. • charge. • energy. • momentum. • potential.
The electron volt is a unit of • capacitance. • charge. • energy. • momentum. • potential.
Two parallel horizontal plates are spaced 0.40 cm apart in air. You introduce an oil droplet of mass 4.9 × 10–17 kg between the plates. If the droplet carries two electronic charges and if there were no air buoyancy, you could hold the droplet motionless between the plates if you kept the potential difference between them at • 60 V. • 12 V. • 3.0 V. • 0.12 kV. • 6.0 V.
Two parallel horizontal plates are spaced 0.40 cm apart in air. You introduce an oil droplet of mass 4.9 × 10–17 kg between the plates. If the droplet carries two electronic charges and if there were no air buoyancy, you could hold the droplet motionless between the plates if you kept the potential difference between them at • 60 V. • 12 V. • 3.0 V. • 0.12 kV. • 6.0 V.
Two parallel metal plates 5.0 cm apart have a potential difference between them of 75 V. The electric force on a positive charge of 3.2 × 10–19 C at a point midway between the plates is approximately • 4.8 × 10–18 N. • 2.4 × 10–17 N. • 1.6 × 10–18 N. • 4.8 × 10–16 N. • 9.6 × 10–17 N.
Two parallel metal plates 5.0 cm apart have a potential difference between them of 75 V. The electric force on a positive charge of 3.2 × 10–19 C at a point midway between the plates is approximately • 4.8 × 10–18 N. • 2.4 × 10–17 N. • 1.6 × 10–18 N. • 4.8 × 10–16 N. • 9.6 × 10–17 N.
The electrostatic potential as a function of distance along a certain line in space is shown in graph (1). Which of the curves in graph (2) is most likely to represent the electric field as a function of distance along the same line?
The electrostatic potential as a function of distance along a certain line in space is shown in graph (1). Which of the curves in graph (2) is most likely to represent the electric field as a function of distance along the same line?
Which of the points shown in the diagram are at the same potential? • 2 and 5 • 2, 3, and 5 • 1 and 4 • 1 and 5 • 2 and 4
Which of the points shown in the diagram are at the same potential? • 2 and 5 • 2, 3, and 5 • 1 and 4 • 1 and 5 • 2 and 4
Which point in the electric field in the diagram is at the highest potential? • 1 • 2 • 3 • 4 • 5
Which point in the electric field in the diagram is at the highest potential? • 1 • 2 • 3 • 4 • 5
Which point in the electric field in the diagram is at the lowest potential? • 1 • 2 • 3 • 4 • 5
Which point in the electric field in the diagram is at the lowest potential? • 1 • 2 • 3 • 4 • 5
The figure shows two plates A and B. Plate A has a potential of 0 V and plate B a potential of 100 V. The dotted lines represent equipotential lines of 25, 50, and 75 V. A positive test charge of 1.6 × 10–19 C at point x is transferred to point z. The electric potential energy gained or lost by the test charge is • 8 × 10–18 J, gained. • 8 × 10–18 J, lost. • 24 × 10–18 J, gained. • 24 × 10–8 J, lost. • 40 × 10–8 J, gained.
The figure shows two plates A and B. Plate A has a potential of 0 V and plate B a potential of 100 V. The dotted lines represent equipotential lines of 25, 50, and 75 V. A positive test charge of 1.6 × 10–19 C at point x is transferred to point z. The electric potential energy gained or lost by the test charge is • 8 × 10–18 J, gained. • 8 × 10–18 J, lost. • 24 × 10–18 J, gained. • 24 × 10–8 J, lost. • 40 × 10–8 J, gained.
Chapter 23: Electric Potential Section 23-2: Potential Due to a System of Point Charges
Charges +Q and –Q are arranged at the corners of a square as shown. When the magnitude of the electric field E and the electric potential V are determined at P, the center of the square, we find that • E≠ 0 and V > 0. • E = 0 and V = 0. • E = 0 and V > 0. • E≠ 0 and V < 0. • None of these is correct.
Charges +Q and –Q are arranged at the corners of a square as shown. When the magnitude of the electric field E and the electric potential V are determined at P, the center of the square, we find that • E≠ 0 and V > 0. • E = 0 and V = 0. • E = 0 and V > 0. • E≠ 0 and V < 0. • None of these is correct.
Two equal positive charges are placed in an external electric field. The equipotential lines shown are at 100 V intervals. The potential for line c is • 100 V. • 100 V. • 200 V. • 200 V. • zero
Two equal positive charges are placed in an external electric field. The equipotential lines shown are at 100 V intervals. The potential for line c is • 100 V. • 100 V. • 200 V. • 200 V. • zero
Two equal positive charges are placed in an external electric field. The equipotential lines shown are at 100 V intervals. The work required to move a third charge, q = e, from the 100 V line to b is • 100 eV. • 100 eV. • 200 eV. • 200 eV. • zero
Two equal positive charges are placed in an external electric field. The equipotential lines shown are at 100 V intervals. The work required to move a third charge, q = e, from the 100 V line to b is • 100 eV. • 100 eV. • 200 eV. • 200 eV. • zero
The potential at a point due to a unit positive point charge is found to be V. If the distance between the charge and the point is tripled, the potential becomes • V/3. • 3V. • V/9. • 9V. • 1/V 2 .
The potential at a point due to a unit positive point charge is found to be V. If the distance between the charge and the point is tripled, the potential becomes • V/3. • 3V. • V/9. • 9V. • 1/V 2 .
The electric field for a charge distribution is E = 0 for r < 1 m, and for . Use the reference point V = 0 as r infinity. The potential for r < 1 m is • 4000 V. • 2000 V. • 1000 V. • Zero. • Cannot be determined precisely.
The electric field for a charge distribution is E = 0 for r < 1 m, and for . Use the reference point V = 0 as r infinity. The potential for r < 1 m is • 4000 V. • 2000 V. • 1000 V. • Zero. • Cannot be determined precisely.
The electric field for a charge distribution is E = 0 for r < 1 m, and for . Use the reference point V = 0 as r infinity. The work required to move a charge, q = e from infinity to r = 2 m is • 4000 eV. • 2000 eV. • 1000 eV. • −4000 eV. • Zero
The electric field for a charge distribution is E = 0 for r < 1 m, and for . Use the reference point V = 0 as r infinity. The work required to move a charge, q = e from infinity to r = 2 m is • 4000 eV. • 2000 eV. • 1000 eV. • −4000 eV. • Zero
Chapter 23: Electric Potential Section 23-3: Computing the Electric Field from the Potential, and Concept Checks 23-1 and 23-2
In what direction can you move relative to an electric field so that the electric potential does not change? • parallel to the electric field • perpendicular to the electric field
In what direction can you move relative to an electric field so that the electric potential does not change? • parallel to the electric field • perpendicular to the electric field
In what direction can you move relative to an electric field so that the electric potential increases at the greatest rate? • in the direction of the electric field • opposite to the direction of the electric field • perpendicular to the electric field
In what direction can you move relative to an electric field so that the electric potential increases at the greatest rate? • in the direction of the electric field • opposite to the direction of the electric field • perpendicular to the electric field
The figure depicts a uniform electric field. Along which direction is there no change in the electric potential?
The figure depicts a uniform electric field. Along which direction is there no change in the electric potential?
The figure depicts a uniform electric field. Along which direction is the increase in the electric potential a maximum?