Recall that the concept of conservative forces, work, and potential energy is useful in solving mechanics problems. It is often possible to solve problems by applying conservation of energy, and arrive at the solution faster and easier than by solving for the motion. The same idea holds with the electrostatic force. In this chapter we will learn about electrical energy, and apply it to a common electrical device called a capacitor.
Recall that the gravitational force, Fg = G m M / r², is a conservative force because the work done to move a particle from point A to point B in a gravitational field depends only on the locations A and B, but not on the path taken from A to B. Because the electrostatic force has the same form as the gravitational force, the electrostatic force is also conservative. For a conservative force, we can define a potential energy function.
Recall that we defined two potential energy functions for the gravitational force: one for the case of a uniform gravitational field, like near the surface of the earth, DPE = mgh; and one for the case of a spherical mass, like a planet, DPE = -G M m / r. (Potential energy and work are measured in joules (J), as is energy.) We will do likewise for the electrical force.
First consider the case of a uniform electrical field, E. Imagine that we move a charge q from point A to B, a distance d parallel to the field. (Provide a drawing.) The force on the charge is F = qE, and is directed parallel to the field. The work done is W = F cosq s (Equation 5.1), where q is the angle between the force and the direction of displacement. The charge is moved a distance d parallel to the force, so we have:
In electric circuits we will frequently be dealing with varying amounts of charge. For this reason, it is even more useful to define something we call the electric potential:
A couple of words about units. Since V = DPE / q = -qEd / q = -Ed (in the case of uniform field). This can also be written as E = -V/d. Therefore we can write the following relation between units: 1N/C = 1V/m. Electric fields are often given the units of V/m.
A uniform electric field of magnitude 250 V/m is directed in the positive x direction. A +12µC charge moves from the origin to the point (x,y) = (20cm, 50cm). (a) What was the change in the potential energy of this charge? (b) Through what potential difference did the charge move?
Begin by drawing a picture of the situation, including the direction of the electric field, and the start and end point of the motion.
(a) The change potential energy is given by the charge times the field times the distance moved parallel to the field. Although the charge moves 50cm in the y direction, the y direction is perpendicular to the field. Only the 20cm moved parallel to the field in the x direction matters for determinig the change of potential energy. DPE = -qEd = -(+12µC)(250 V/m)(0.20 m) = -6.0×10-4 CV = -6.0×10-4 J.
(b) The potential difference IS the difference of electric potential, DV. DV = DPE / q = -6.0×10-4 J / 12µC = -50 V.
(a) Through what potential difference would an electron need to accelerate to achieve a speed of 60% of the speed of light, starting from rest? (The speed of light is 3.00×108 m/s.)
(a) The final speed of the electron is vf = 0.6(3.00×108 m/s) = 1.80×108 m/s.
At this speed, the energy (non-relativistic) is the kinetic energy,
E = ½mvf² = 0.5(9.11×10-31 kg)(1.80×108 m/s)² = 1.48×10-14 J.
This energy must equal the change in potential energy from moving through a potential difference, V, E = DPE = qV.
Therefore:
V = E/q = (1.48×10-14 J)/(-1.6×10-19 C) = -9.25×104 V.
The electric potential a distance r from a point charge q is given by:
The potential energy required to bring a charge q2 from infinity to a distance r from charge q1 is given by
Three charges are situated at corners of a rectangle as in Figure P16.13. How much energy would be expended in moving the 8.0µC charge to infinity?
(Drawing) We are dealing with point charges, so we will take the zero of potential and potential energy at infinity. Let's solve the problem using W = DPE = q(VB - VA), where q is the 8.0µC charge, A is the upper left corner of the rectangle, and B is infinity. (Note that it doesn't matter which direction we move away to, as long as it is infinitely far away from the charges.)
As noted above, VB = 0.
Let's determine VA.
It has two contributions, one from the 2.0µC charge, V2, and one from the 4.0µC charge, V4.
Recall that the prefix µ means 10-6.
V2 = k q2 / r2 = (9×109)(2.0µC)/(0.030 m) = 6.0×105 V.
V4 = k q4 / r4 = (9×109)(4.0µC)/sqrt{(0.030 m)² + (0.060 m)²} = 5.4×105 V.
Summing these 2 contributions gives a total potential:
VA = 11.4×105 V.
The energy expended is:
W = (8.0µC)(0 V - 11.4×105 V) = -91×10-1 CV = -9.1 J.