Download Electric Fields and Forces Between Charges: Discussion Questions and more Exercises Physics in PDF only on Docsity!
Discussion Question 2A P212, Week 2 Electric Field Lines Electric fields are force fields produced by electric charges. They can extend throughout space and influence other charges (and currents and magnetic fields, as you will discover later in the course). Electric Field Lines graphically represent electric fields. They show you the direction and relative magnitude of force that a test charge would feel in the field region. (a) The configuration at right shows some points around a positive charge Q. If a positive test charge q is placed at point A, at a distance r A from Q , calculate the magnitude of the force on the test charge in terms the given variables.
|๐น๐ด| = kqQ/๐๐ด^2
Draw the direction of the force with an arrow on the diagram. (b) Would the magnitude of the force be bigger or smaller if the charge was placed at points B,C, or D? Force would be bigger at B, smaller at C or D Draw the direction of these forces on the diagram with arrows, using the length of the arrows to indicate the relative magnitude of the forces. (c) On the diagram at right, draw extended electric field lines for Q. What feature of these lines indicates the relative magnitude of the fields at A and B? Electric fieldsโand hence magnitudes of forces on test chargesโare strongest where field lines are most dense. So, it is clear from the field lines that for test charges placed at the different points, |FB| > |FA| > |FC| > |FD|. (d) Calculate the magnitude of the electric field at point A, in terms of given variables.
๐ธโ๐ด = ๐๐/๐๐ด^2 ๐ฬ
If Q is doubled, what is the new value of the electric field at A? If q is doubled instead, what is the new value of the field at A? If Q is doubled, E ๏ข = 2 E
If q is doubled E^ ๏ข = E (e) The region in the box at right contains a uniform electric field, E. A test charge q is placed in the box. Compare the magnitude of the field at points A and B. Write an expression for the force the test charge feels at point A.
(f) The region in the box at right contains a non-uniform electric field. Compare the magnitudes of the field at A,B, and C. If test charges of equal magnitudes (and masses) are placed at B and C, which one is pushed out of the box first?
The charge at B is pushed out of the box first.
E^ โข^ A^
โข B
E
โข A
B
C
Discussion Question 2C P212, Week 2 Forces Between Point Charges Consider two charges, Q , placed at fixed positions a horizontal distance, d , apart as shown in the figure. A third particle of mass m and unknown charge q is then placed a vertical distance h above the fixed charges. The position of this third particle is arranged so that it is suspended in mid-air above the other two! This equilibrium condition occurs because the electrical and gravitational forces on the charge precisely cancel each other. (a) Calculate the charge q that is required to balance the third particle in mid-air. Express your answer in terms of the given parameters plus physical constants. 2 ๐๐๐ (๐/ 2 )^2 + โ^2 โ โ(๐/^2 )^2 +^ โ^2 โ ๐๐ = 0 ๐ = ๐๐ 2 ๐พ๐โ [(๐/ 2 )^2 + โ^2 ]^3 /^2 (b) Does your answer make sense? Find three limiting cases where you know what the answer should be and make sure your expression for q gives the right result. And hereโs another important check: does the sign of q make sense? Want q and Q to have same sign to insure repulsion. As hโ0: qโ๏ฅ โ
As d โ0: ๐ =
๐( 2 ๐)๐ โ^2
๐๐โ^2 2 ๐๐ โ As mโ0: qโ 0
Q Q
h
d
q, m
(c) Equilibrium situations like the one you have analyzed can be described as either stable or unstable. Stable equilibrium means that the system will return to its equilibrium configuration when disturbed by a small amount (like a swinging pendulum). Unstable equilibrium describes situations like a ball balanced on top of a hill โ the slightest disturbance will drastically affect the system. Think about your suspended charge problem for a moment ... what sort of equilibrium condition is present? Take your best guess before continuing. No restoring force in ๏ฑ z (in-out of page) thus unstable equilibrium Stable equilibrium means that if you โbumpโ the charge, the system will provide a restoring force to bring it back to its resting place. In the opposite case of unstable equilibrium, the system will encourage the displacement, forcing the charge further away from its equilibrium spot. So letโs try bumping our suspended charge in different directions โฆ Note: for the graphs below, we are taking the equilibrium height h of the charge to be 1 meter.^1 (d) Is the balanced charge stable against displacements in y? If you bump the charge in the ๏ฑ y direction (up or down), will the combined forces of gravity and electricity provide a restoring force, or will they push it further away? Check your physical intuition, then consult the plot at right which shows the vertical component Fy of the force as a function of y. (Note how the force goes to zero at the equilibrium spot y = h = 1 m, as it must โฆ but here weโre concerned with the behavior of Fy around this point.) Near ๐ฆ = 1 , the force goes negative as you increase in y and positive as you decrease in y. This has the effect of pushing the charge back to the equilibrium point. (^1) For the record, the other parameters used to produce these plots are d = 2 m, m = 3 g, Q = +2.5 ๏ญC, and q = 1.85 ๏ญC. x y
h
Q Q
q, m
Wrapup: If all went well on the previous page, you found that your suspended charge is stable against displacements in the x and y directions, but not in z. Itโs challenging to think in 3 dimensions! We worked with force plots on the previous page, but the potential energy map U ( x , y , z ) of a system is a much better way to visualize its behavior. Remember from Physics 211 how a system always wants to minimize its potential energy? That means that wells in the potential energy distribution indicate places of stable equilibrium โ places where the system wants to be. Below are three plots showing the map U ( x , y , z ) for your suspended charge (youโll find out how to calculate them next week). Notice how clear the conditions of stable and unstable equilibrium are!
Discussion Question 2D P212, Week 2 An Electrostatic Quadrupole The figure below shows a head-on view of four long, parallel charged rails (they extend both into and out of the page). The side rails carry a positive charge per unit length +๏ฌ. The top and bottom rails have the same charge density, but with opposite sign (i.e. they are negatively charged). This arrangement forms an electrostatic quadrupole. In the questions below, you will discover why it is useful in devices which contain beams of moving charged particles. We will not perform any calculations in this question โฆ all you need is your physical intuition. (a) Sketch the electric field lines in the interior region of the quadrupole. Lines go from positive to negative poles. (b) Given your sketch, where is the electric field the strongest? Where is it the weakest? Strongest at any pole, weakest at center. (c) Now imagine that a beam of moving electrons enters the quadrupole, with velocity parallel to the charged rods. When the beam enters the device, its cross-sectional profile is circular. What will the beam look like when it exits the quadrupole on the other side? (Assume that the force exerted on the electrons is not too large.) The beam of electrons would be flattened into an oval that is long in the x-axis and short in the y-axis. (d) The cathode ray tube (CRT) in a TV monitor contains a beam of electrons directed towards a fluorescent screen. The screen lights up where the beam hits it. If you were an electrical engineer designing such a CRT, why might you find it useful to pass the electron beam through a quadrupole before it hits the screen? (A little hint: quadrupoles are usually used in pairs โฆ can you figure out why?) The quadrupole acts to correct errors in the electron firing gun. Non-perfect electrons will still hit pixel. x y + ๏ฌ + ๏ฌ
- ๏ฌ - ๏ฌ beam of electrons heading into page y x
๏ฌ
๏ฌ
- ๏ฌ - ๏ฌ rods extend into and out of page d d
