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Charged particles have energy due to their positions. A group of positively charged particles squeezed tightly together has just such energy. When released they will fly apart, converting the potential energy into kinetic. Two charges, Q and q separated by a distance r have electric potential energy

U = kQq/r where k = 9 x 10^{9} . When r = infinity, U = 0.

Electric potential energy is the work required to bring charges from infinite separation to separation r.

The charge , Q, causes every other charge, q, to have electric potential energy. Q is called the source charge as it is considered to be the cause of the electric potential energy, while q is called a test charge.

A common definition for electric potential is the work done per unit charge moving a test charge from infinity to a point in an electric field. The units are J/C, or volts (V).

V = U/q

Therefore, U = Vq

That is, the work done on every coulomb of charge of a test charge to bring it from infinity to at a distance r away from the source of the field is

V = kQ/r

The work done moving a charge, q, from one point with potential V_{1} to another point with potential V_{2} in the field is

U = q(V_{2} -V_{1})

Consider a charged particle in an electric field. If the particle had twice the charge, which of the following statements would be true?

a. The electric potential energy and the electric potential would both be twice as great.

b. The electric potential energy would be twice as great and the electric potential would be the same.

c. The electric potential energy would be the same and the electric potential would be twice as great.

d. The electric potential energy and the electric potential would both be the same.

b. The electric potential energy would be twice as great and the electric potential would be the same.

(See bottom of page for answers.)

1. Two protons move toward each other at 4 X 10^{6} m/s. How close together do they get?

2. Three charges, q_{1} = 4 x 10^{-6} C, q_{2} = -2 x 10^{-6} C, and q_{3} = 5 x 10^{-6} C are placed at the corners of an isosceles triangle with sides 0.30 m. What is the potential energy of the system?

3. What is the speed of an electron that has been accelerated from rest through a potential difference of 80.0 kV?

4. A positive charge of magnitude, q, is placed at position, A, between two charged parallel plates as shown to the left. The parallel plates have a potential difference of ΔV and are separated by a distance, d. State the increase or decrease in potential energy of the particle when it is moved from (a) A to B; (b) B to C; (c) C to D; (d) D to E

5.

a. A particle with a charge of -3 X 10^{-8} C and 2 X 10^{-5} J of kinetic energy enters the region between two charged parallel plates as shown to the left. How far into the region will the particle travel?

b. If an electron were released from rest next to the negative plate, with what speed would it strike the positive plate?

6. Through what potential difference would an electron have to be accelerated to give it a deBroglie wavelength of 10^{-10} m? (velocity electron<<velocity light).

7. A particle (charge 7.5 x 10^{-6} C) is released from rest 0.1 m away from a fixed 2.0 x 10^{-6} C charge. What is the kinetic energy of the particle 1.0 m away from the fixed charge?

8. A 0.08 nC is located at x = 0.05 m. What is the electric potential (relative to zero at infinity) at a point x=0.18m, on the x-axis?

9. Three point charges, each with a charge of 1.50 µC, form an equilateral triangle with 0.200 m sides. What is the potential energy of the system?

10. Two protons and an alpha particle form an equilateral triangle with sides 9.20×10^{-10} m. The particles are released. What is the total energy of the particles when they are far apart?

11. An electron moving to the right at 1.0% the speed of light comes to rest after moving 4.0 cm through a uniform electric field parallel to its direction of motion. What is the direction and the strength of the electric field?

12. Consider two 2.75 µC charges. One charge is fixed. The other resides in 3.30 g particle released from rest 1.15 m east and 0.490 m north of the fixed charge. What is the maximum speed the particle will attain? At what distance from the fixed charge will the particle have half of its maximum speed?

13. What is the charge of q if location A is 2.40 m from q, location B is 4.5m q, and the electric potential difference between points A and B is 45V?

14. A 4 nC point charge is 39 cm from a 5nC point charge. What is the electric potential of a point midway between the charges?

15. An electron that is initially 53 cm away from a proton is moved so that the change in electric potential energy as a result of this movement is 3 x 10^{-28} J. What distance finally separates the electron and the proton?

16. If 4.8x10^{-6} J of energy is used to move a charge through an electric potential
difference of 2.0x10^{-6} V, what is the
magnitude of the charge?

Selected solutions are printed below.

For solutions to all the problems on this page click here.

1. When no outside forces such as friction acts on an system, the total energy of the system is constant. That is, the sum of kinetic, and potential energies at one point is the same as at any other time.

E_{total} = (0.5)mv^{2} + U is unchanging if no work is done on the system.

At infinite separation and since the system contains 2 protons, the total energy is

E_{total} = 2(0.5)mv^{2} + kQq/r = 2(0.5)(1.67 X 10^{-27} kg)(4 X 10^{6} m/s)^{2} + 0

= 2.67 X 10^{-14} J

At their closest approach the two positively charged particles have slowed to a stop:

E_{total} = 2(0.5)mv^{2} + kQq/r

2.67 X 10^{-14} J = 0 + (9 x 10^{9})(1.6 X 10^{-19} C)(1.6 X 10^{-19} C)/r

r = 8.63 X 10^{-15} m

For solutions to all the problems on this page click here.

2. The potential energy of the system is the sum of the potential energies of each pair of charges:

U = U_{1,2} + U_{1,3} + U_{2,3}

U_{1,2} = (9 x 10^{9})(4 x 10^{-6} C)(-2 x 10^{-6} C)/(0.30 m) = -0.240 J

U_{1,3} = (9 x 10^{9})(4 x 10^{-6} C)(5 x 10^{-6} C)/(0.424 m) = 0.425 J

U_{2,3} = (9 x 10^{9})(-2 x 10^{-6} C)(5 x 10^{-6} C)/(0.30 m) = -0.300 J

U = -0.115 J

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