Chapter 19
A point charge q=-0.35nC is fixed at the origin. Where must an electron be placed in order for the electric force acting on it to be exactly opposite to its weight?
0.002375m above point q (F = k*q*e/r^2, mg = k*q*e/r^2, r = sqrt(k*q*e/mg^2), with e = 1.6 e -19 C and m = 9.11e-31 kg)
1 Electron = __ Coulombs
1.60 e -19 Coulombs
Two charges, Q1 and Q2, are separated by a certain distance. The ratio of charge Q1 to Q2 is 1:2. The ratio of force on Q1 to the force on Q2 is....
1:1 (b/c force between two objects is the same due to gravity being only attractive)
Three equal point charges of varying signs are placed on the corners of a square of side d as shown in Figure 19-10. Which of the arrows shown represents the direction of the net electric field at the center of the square?
A, or whichever arrow is pointing in the direction of the corner with no charge.
Four point charges of varying magnitude and size are arranged on the corners of the square of side d as shown in Figure 19-8. Which of the arrows shown represents the net force acting on the point charge with a charge +Q? (The charge on the corner opposite to charge +Q is charge -Q. The other charges, situated on the bottom left and top right corner are +2Q.)
A, the arrow pointing to the immediate Northwest and lying outside of the square.
++++++++++++++++++ ------------------------------- Which of the arrows shown in Figure 19-14 represents the correct direction of the electric field between the two metal plates? (The one pointing down or up?)
A, the one pointing down. (Remember that direction of an electric field is the direction of the force experienced by a positive charge.)
Four point charges of equal magnitude and signs are arranged on three of the corners and at the center of the square of side d as shown in figure 19-4. Which of the arrows shown represents the net force acting on the center charge? (All charges are -Q)
B, pointing towards the immediate Northeast corner.
Electrical and gravitational forces follow similar equations with one main difference:
Gravitational forces are always attractive but electrical forces can be attractive OR repulsive (think: we always fall to earth, never upwards)
Two charges, Q1 and Q2 are separated a certain distance R. If the magnitudes of the charges are doubled and their separation is also doubled, then what happens to the electrical force between these two charges?
It remains the same.
Three equal point charges of varying signs are placed on the corners of a square of side d as shown in Figure 19-10. What is the correct expression for the magnitude of the net electric field at the center of the square?
SAY IT: E EQUALS TWO Q OVER D SQUARED: E =k(2q/d^2) (Consider that only two charges actually have any effect on the center charge, thus making the coefficient 2 instead of 3.)
A system of 1525 particles...has a net charge of -5.456e-17 C....
a) 933 electrons (Net charge/# of coulombs in 1 electron) = -answer, 1525+answer= secondAnswer, secondAnswer/2 = finalAnswer b) 9.89 e -25 kg
Two point charges lie on the x axis. A charge of 6.2 uC is at the origin, and a charge of -9.5 uC is at x= 10 cm. What is the net electric field at a) x=-4 cm and at b) x=+4 cm?
a) subtract two electric charges USE CORRECT LENGTHS b) add two electric charges USE CORRECT LENGTHS
When the distance between the two charges is doubled, the force between them is...
reduced by a factor of 4. (Force is inversely related to distance and distance is squared.)
The charge of the proton is the same as...
the charge on the electron, but of the opposite sign.
The number of protons in the nucleus of an electronically neutral atom is equal to...
the number of electrons surrounding the nucleus. (neutral = same amount of positive as negative)
An electron and a proton are released simultaneously from rest and start to move towards each other because of the attractive Coulomb force between them. They are initially separated by a distance d. The two particles eventually collide. When they collide...
they are closer to the proton's initial position. (Remember that the force between the electron and proton is the same, but electrons have less mass; thus a=F/m (from F=ma) proves that electrons move faster, and will approach the protons initial position quicker than the proton can approach the electrons initial position.)