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Real Insulators (Dielectrics). If I bring a charged rod to a leaf electrometer: A] nothing will happen B] nothing will happen until I touch the electrometer with the rod - then the leaves will spread apart C] the leaves will spread apart when I get the rod close to the electrometer.
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If I bring a charged rod to a leaf electrometer:A] nothing will happenB] nothing will happen until I touch the electrometer with the rod - then the leaves will spread apartC] the leaves will spread apart when I get the rod close to the electrometer
What is the direction of the electric force on a negative charge at 1? A] to the rightB] to the leftC] upD] down What is the direction of the electric force on a positive charge at 2?
If I put a + charge at 3., what is the direction of the electric force?A] straight down, directly toward the - chargeB] straight up,directly away from the - chargeC] a little to the right from straight downD] a little to the left from straight down
What could be the charges here?Assume that all field lines are shown. A] Q1 = +3C, Q2 = -2 CB] Q1 = +18C, Q2= -10 CC] Q1= -18C, Q2=+10 CD] Q1 = +9 C, Q2 = +5 CE] none of these
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Electric field is defined by the force on a “test charge”. Since forces add by superposition, fields do also. • Field lines never cross each other • Field lines are perpendicular to conductors at their surfaces • Field lines share the symmetry of the sources & sinks • Density of field lines (in 3D) is proportional to field strength Field lines:
Which picture shows the field lines from an infinite sheet containing a uniformly dense negative charge?
What is wrong with picture B? A] field lines are symmetric top/bottom B] field lines cross C] field lines are symmetric right/left D] field lines should start at sheet
What’s wrong with Figure C? A] it has the wrong symmetry B] it has the wrong direction of field lines C] field lines have sharp turns
Two source charges have equal magnitude. One is + and the other -, as shown. What vector could represent the electric field at the point shown caused BY THE POSITIVE CHARGE ALONE? Given that B is the field from the + charge alone, what vector could be the total field from both charges?
Calculus: how do we find E fields from extended charge distributions? Use symmetry Apply Coulomb’s Law to infinitesimal parts Take components Use calculus to sum up all contributions for either the x or the y component
We draw E-field vectors with their tails at the “point of interest” i.e. the point in space at which the E-field is shown. We draw E-field vectors on pictures showing the locations of charges, but the E-field vector is NOT a displacement vector, of course. It has units of Force per charge, or Newtons per Coulomb. E field vectors are everywhere tangent to E field lines. On a drawing of E field lines, there is a field in between the lines. The lines are just a way of illustrating the vector E field. The strength of the E field is proportional to the density of field lines (number of lines per unit perpendicular area.)
Could the E field shown be caused by a single point charge? A] no B] yes
Could the E field shown be caused by a single point charge? A] no B] yes Field lines are not the same as field vectors. Field lines are tangent to field vectors (at their tails, of course!)
Could the E field shown be caused by a single point charge? A] no B] yes
Could the E field shown be caused by a single point charge? A] no B] yes
Could the E field shown be caused by a single point charge? A] no B] yes
This field CAN be caused by a single point charge. Where must it be? (or choose E: either not shown, or it cannot be determined.)
This is an exercise in reasoning about proportions. At point D, the field is four times as strong as at the rightmost tail. How much closer is D to the source than the rightmost tail? A] D must be half as far from the source B] D must be one quarter as far from the source C] D must be one eighth as far from the source
Two charges at rest act on each other by the Coulomb force. Could these vectors represent the forces on each particle? A] yes B] no Although the forces are equal and opposite, they are not collinear. From coulomb’s law, the forces act along the line between the particles. Coulomb’s law conserves momentum (because forces are equal and opposite) And angular momentum (because forces are collinear.) We will see with MOVING charges, electromagnetic forces do NOT conserve either momentum or angular momentum (of the particles.) As we are very reluctant to abandon these conservation laws, we believe that EM fields can carry momentum and angular momentum.
A] + B] - C] can’t tell
A] + B] - C] can’t tell The left point is closer to the charge, because the field is stronger there. The field points away from the charge, so it is +
X Y We reason that the charge is to the left, and +. What is the ratio of the distance of the charge from point Y to the distance of the charge from point X? A] 4/9 B] 2/3 C] 3/2 D] 9/4 E] There’s no way to know.
A B C D X Y Because the field varies like 1/r2, the distances must be in the ratio 2:3. Y is farther (because the field is weaker) so the ratio of distance to Y to distance to X is 3/2. Where is the point charge? (Choose A for any point to the left of B)