1) The potential at location A is 441 V. A positively charged particle is released there from rest and arrives at location B with a speed vB. The potential at location C is 805 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2 vB. Find the potential at B.
for this, i place the drawing as
A B C
and i am assuming that A + C = B, i am not sure if that's correct or not and i just added the 2 different potential. But then the Vb and 2Vb isnt used and I am not sure where its used unless its telling me that the distance between A and B is half of B and C.
2) Four identical charges (+1.6 ìC each) are brought from infinity and fixed to a straight line. Each charge is 0.36 m from the next. Determine the electric potential energy of this group.
i placed the charges as follow
q1 q2 q 3 q4
and used q1 as the starting point so EPE = 0
then i used the equation V = Kq/r and multiple the V by (1.6 micro columb) to get the EPE. I did that for q2 q3 q4 and changing r as i get closer to q4. Then i added all the EPE together but i am missing something.
EPE of q1 = 0
EPE of q2
(q1)(k)/(.36m) = V
EPE2 = (V)(1.6 micro columb)
EPE of q3
(q1)(k)/(.36m x 2) = V
EPE3 = (V)(1.6 micro columb)
EPE of q4
(q1)(k)/(.36m x 4) = V
EPE4 = (V)(1.6 micro columb)
Add EPE (1 + 2 + 3 + 4) = Total EPE of the group
but neither is correct i am doing stuff wrong, please help. thank you
2) Your EPE of particles 1 and 2 are OK. For particle 3, you need to compute the work needed to approach BOTH the q1 and q2 particles already present. For q4, you need to consider the interaction with q1, q2 and q3. The separation distances will differ in the last two cases.
i change the equation and got thsi
i placed the charges as follow
q1 q2 q 3 q4
and used q1 as the starting point so EPE = 0
then i used the equation V = Kq/r and multiple the V by (1.6 micro columb) to get the EPE. I did that for q2 q3 q4 and changing r as i get closer to q4. Then i added all the EPE together but i am missing something.
EPE of q1 = 0
EPE of q2
(q1)(k)/(.36m) = V
EPE2 = (V)(1.6 micro columb)
EPE of q3
(q)(k)/(.36m) + (q)(k)/(.36m x 2) = V
EPE3 = (V)(1.6 micro columb)
EPE of q4
(q)(k)/(.36m) + (q)(k)/(.36m x 2) + (q)(k)/(.36m x 4) = V
EPE4 = (V)(1.6 micro columb)
Add EPE (1 + 2 + 3 + 4) = Total EPE of the group
For question 1, let me walk you through the solution.
First, let's assume the potential at B is Vb.
From the given information, we know that the particle starts at A with a potential of 441 V and arrives at B with a speed of vB. We can use the principle of conservation of mechanical energy to relate the initial potential energy at A to the final kinetic energy at B.
The initial potential energy at A is given by qV, where q is the charge of the particle. Since the particle starts from rest, all the initial energy is potential energy. So, the initial potential energy at A is q(441 V).
The final kinetic energy at B is given by (1/2)mvB^2, where m is the mass of the particle. We don't know the value of m, but it cancels out when we compare the initial and final energies.
Now, let's consider the situation at C. The potential at C is 805 V. When the particle is released from rest at C, it arrives at B with twice the speed it previously had, or 2vB. Again, we can use the principle of conservation of mechanical energy.
The initial potential energy at C is q(805 V), and the final kinetic energy at B is (1/2)m(2vB)^2 = 2mvB^2.
Since the initial and final energies must be equal, we can set up the following equation:
q(441 V) = 2q(805 V) - 2mvB^2
Notice that the mass cancels out, which means the mass of the particle is not necessary for the solution.
From this equation, we can solve for vB, which is the speed of the particle at B. Once we have vB, we can find the potential at B by using the formula for kinetic energy: (1/2)mvB^2.
Now, let's move on to question 2.
Your approach for finding the electric potential energy (EPE) of each charge is correct. However, there is an additional consideration for particles 3 and 4.
For particle 3, you need to compute the work needed to approach BOTH the q1 and q2 particles already present. This means you need to consider the effect of electric potential from both q1 and q2 when calculating the potential energy of particle 3.
The equation for particle 3's EPE would be:
EPE3 = (V1)(1.6 μC) + (V2)(1.6 μC)
Where V1 and V2 are the electric potential due to q1 and q2 respectively. The values of V1 and V2 can be calculated using the formula V = kq/r, where k is the electrostatic constant, q is the charge, and r is the distance.
Similarly, for particle 4, you need to consider the interaction with q1, q2, and q3. The equation for particle 4's EPE would be:
EPE4 = (V1)(1.6 μC) + (V2)(1.6 μC) + (V3)(1.6 μC)
Again, calculate V1, V2, and V3 using the formula V = kq/r.
Finally, add up the EPE of all the particles: (EPE1 + EPE2 + EPE3 + EPE4), and you will get the total electric potential energy of the group of charges.