Download Honors Physics Electric Potential Energy and Potential Difference

Honors Physics Electric Potential Energy and Potential Difference Worksheet
1. Potential of Two Point Charges Problem
Two point charges q1 = -88.0 nC & q2 = 360 nC are located at the fixed positions at x1 = 22.0 cm and x2 = 96.0 cm on
the x-axis.
(A) What is the electric potential due to both charges at x = 50.0 cm?
(B) Where on the x-axis is the electric potential due to both charges equal to zero?
(C) How much work is required to move a third charge q3 =13.0 mC from the origin to x = 50.0 cm along any path (that
does not go through one of the charges)?
(D) If a fourth charge q4 with a charge of -11.0 mC and a mass of 3.90 gm is released from rest at the origin, what is the
charge's final speed and direction when it is very far away from the other two charges?
2. A proton is accelerated from rest through a potential of 500 volts. What is its final kinetic energy?
3. What is the final velocity of the proton in the previous problem? (The mass of the proton is 1.67 x 10–27 kg.)
4. If Q1 and Q2 are equal and oppositely charged, what is the
potential at point B?
5. If Q1 is twice Q2 and both are positive, where can a point of
zero potential be found?
6. An electric field is created by a Van de Graaf generator. A test charge of (q) is placed in the field at a distance r.
a) If the test charge has a magnitude of +1q write an expression for
i) the potential energy between the charges
ii) the potential of the system
b) if the test charge has a magnitude of +2q write an expression for
i) the potential energy between the charges
ii) the potential of the system
7. The potential difference between a storm cloud and the ground is 100-million volts. If a charge of 2.0 C transfers (via a
lightning bolt), how much potential energy was transferred?
8. A system involves a +6μC charge and a +4μC charge initially separated by an infinite distance.
a) How much work is required to bring them 150 cm from each other?
b) What is the total voltage at a point in the middle of the system?
9) For the uniform field shown on the right, an electron is released from rest from plate B. Determine the:
a) work done by the field
b) force on the charge
+
c) acceleration of the charge
+
d) the speed of the charge when it reaches plate A
E = 500 N/C
-
+
e) change in potential energy
+
5 mm
A
B
+
10. A charge of -4.0μC is initially 400 mm from a fixed charge or -6.0μC and is then moved to a position 90 cm from the
fixed charge.
a) what is the change in the mutual potential energy of the charges?
b) Does this change depend on the path through which the one charge is moved?
*11. A proton moving directly toward another fixed proton has a speed of 5.0 m/s when the two are 100 cm apart. How
close to the stationary proton will the moving proton be when it stops and reverses course?
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Solutions
(A) Find V(x = .50 m).
Since 50 cm is in between the two charges the electric potential must be written as
1.
if r1 and r2 are to be positive numbers.
(B) Find the points xo on the x-axis were V = 0.
The potential immediately surrounding a positive charge is positive while the potential immediately surrounding a negative charge is negative. Some were in between the two charges the two
potentials must add to up to zero. Since this region is greater than x1 and less than x2, then electric potential of two point charges in between the charges is
if r1 and r2 are to be positive. Setting V(xo) equal to zero and solving for xo,
(C) Find the work to move a 13.0 C from the origin to a point 50.0 cm from the origin on the positive x-axis.
First note that you cannot simply move the charge along the x-axis since potential and electric field are infinity at the location of the two fixed charges. Thus we will have to move the charge
off the x-axis to physically move the charge to x = 50.0 cm .
Approach I: The hard way but very instructive.
We will brake the problem into two parts. First we will find the work due to charge q1 alone. Next we will find the work due to charge q2 alone and added the results.
Work due to charge q1: If we move q3 in a semicircle around q1 back onto the x-axis this process will take no work since the potential due to q1 is constant along this arc, it is an equipotential
line. The charge q3 will then be at x = .440 m from the origin. Now move it away from q1 along the x-axis to 50.0 cm.
The work needed to move q3 from .44 m to .50 m can be calculated from the fundamental definition of work and the Coulomb's Law.
There are two errors in the above solution that have canceled each other out. I did not notice it at first sine the answer came out positive as it should be to pull a positive charge away from a
negative charge. The force F13 and the line integration are in the opposite direction and the F13 should be the magnitude which is a positive number. The correct solution should be
The work due to charge q2:
By removing q1, the work need to move q3 from origin to .50 cm along the x-axis due to q2 alone is,
The sign cannot be negative because itis going to take positive work to move two positive charges closer together. If we wanted to fix up the math to make it come out positive we should
revisit the definition of work more carefully,
The direction of integration (to the right) and the direction of the force on q3 (to the left) are in opposite directions.
Thus the total work is
Approach II: The Easy Way
From the definition of potential,

(D) Find the final velocity of an -11.0 C charge q4 if it is released from rest at the origin and it has a mass of 3.90 gm.
The wording of the question implies that the negative charge q4 will move away from the origin and go to minus infinity. Assuming that this is true let us use the conservation of energy to find
the velocity when the charge reaches infinity. We know that the potential energy of the electric field is just PE = qV and that the potential of a point charge is zero at infinity.
If you run the simulation for this problem with the charge placed at the origin you will see that the charge reaches a speed of 1.27 m/s when it only 130 m from the origin. This is typical
because the potential goes as one over the distance. At 130 m the potential has dropped to from -225 V to18.6 V