Electric Potential: Two identical + 8.5mu C point charges

In summary: I was getting it to be 114 m/s. In summary, two identical + 8.5mu C point charges, initially 2.5 cm apart and with identical masses of 2.0mg, are released from rest and will have a final speed of approximately 720 m/s when they are very far apart from each other. This is found by using the equations ΔKE + ΔPE = 0 and ΔPE = qV, and taking into account that the potential energy will be converted to kinetic energy. The electric potential due to only one of the charges can be found at a distance of 2.5 cm from that charge, and this value can be used to calculate the initial potential energy of the system.
  • #1
PeachBanana
191
0

Homework Statement



Two identical + 8.5mu C point charges are initially 2.5 cm from each other. If they are released at the same instant from rest, how fast will each be moving when they are very far away from each other? Assume they have identical masses of 2.0mg .


Homework Equations



ΔKE + ΔPE = 0
ΔPE = qV


The Attempt at a Solution



I went to a tutoring session today and the tutor told me ΔPE = -kq^2/r. I have no idea from where this derivation came. I'm looking at my book right now and it says ΔPE = qV. Also, she said since there are two balls ΔKE = mv^2 as opposed to 1/2mv^2, which I understand. If I'm going to use ΔPE = qV, I'm unsure how to find the electric potential difference.
 
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  • #2


]
PeachBanana said:

Homework Statement



Two identical + 8.5mu C point charges are initially 2.5 cm from each other. If they are released at the same instant from rest, how fast will each be moving when they are very far away from each other? Assume they have identical masses of 2.0mg .

Homework Equations



ΔKE + ΔPE = 0
ΔPE = qV


The Attempt at a Solution



I went to a tutoring session today and the tutor told me ΔPE = -kq^2/r. I have no idea from where this derivation came. I'm looking at my book right now and it says ΔPE = qV. Also, she said since there are two balls ΔKE = mv^2 as opposed to 1/2mv^2, which I understand. If I'm going to use ΔPE = qV, I'm unsure how to find the electric potential difference.
What is the electric potential due to only one of the charges, at a distance of 2.5 cm from that charge?
That's your V.

Then the initial potential energy of the system of two charges is qV, for that V.​
 
  • #3


These two equations are all you need to solve the problem, keeping in mind that all of the potential energy will be transferred to kinetic energy.

Potential energy of 2 point charges q1 and q2:
[itex]\Large U_{E}=\frac{kq_{1}q_{2}}{r}[/itex]

Kinetic energy:
[itex]\Large K_{E}=\frac{1}{2}mv^{2}[/itex]

Both of these equations are basic equations that you'll find in any physics textbook, by the way.
 
  • #4


Now what I've done is:

U initial = (8.5*10^-6)(8.99*10^9 N*m/C^2) / (0.025 m)^2

U initial = -1.0392*10^3 J

-1.0392*10^3 J +(2)1/2mv^2 = 0

(0.002g) v^2 = 1.0392*10^3 J

v^2 = 5.196*10^5 m^2/s^2

v= 720 m/s

Hmm.
 
  • #5


PeachBanana said:
Now what I've done is:

U initial = (8.5*10^-6)(8.99*10^9 N*m/C^2) / (0.025 m)^2

U initial = -1.0392*10^3 J

-1.0392*10^3 J +(2)1/2mv^2 = 0

(0.002g) v^2 = 1.0392*10^3 J

v^2 = 5.196*10^5 m^2/s^2

v= 720 m/s

Hmm.

Your initial potential should be positive, not negative. The ΔU is negative because it goes from the initial value to 0. Otherwise, it looks alright. Your answer shouldn't change.
 
  • #6


PeachBanana said:
Now what I've done is:

U initial = (8.5*10^-6)(8.99*10^9 N*m/C^2) / (0.025 m)^2

U initial = -1.0392*10^3 J

-1.0392*10^3 J +(2)1/2mv^2 = 0

(0.002g) v^2 = 1.0392*10^3 J

v^2 = 5.196*10^5 m^2/s^2

v= 720 m/s

Hmm.
[itex]\displaystyle U_{\text{initial}}=\frac{k\,q^2}{r}[/itex]

You have [itex]\displaystyle U_{\text{initial}}=\frac{k\,q}{r^2}\,,[/itex] which is incorrect.
 
  • #7


I really hope I'm not doing something silly. Does everyone else get the initial U value to be 25.98 J?

(8.5*10^-6 C)^2 * (8.99*10^9 Nm/C^2) / (0.025 m) = 25.98 J

Then from there get the final answer to be ~ 114 m/s?
 
  • #8


SammyS said:
[itex]\displaystyle U_{\text{initial}}=\frac{k\,q^2}{r}[/itex]
You have [itex]\displaystyle U_{\text{initial}}=\frac{k\,q}{r^2}\,,[/itex] which is incorrect.
Oh right, I missed that. Sorry.

PeachBanana said:
I really hope I'm not doing something silly. Does everyone else get the initial U value to be 25.98 J?
(8.5*10^-6 C)^2 * (8.99*10^9 Nm/C^2) / (0.025 m) = 25.98 J
Then from there get the final answer to be ~ 114 m/s?

Your potential energy (25.98 J) is correct. The final answer seems to be a bit different. Try solving the kinetic energy equation for v again, and double check that you're using the right values for KE and m.
 
  • #9
Found my problem; I was using "g" instead of "kg" for mass.
 

1. What is the electric potential at a point between two identical + 8.5mu C point charges?

The electric potential at a point between two identical + 8.5mu C point charges can be calculated by using the formula V = kq/r, where V is the electric potential, k is the Coulomb's constant, q is the magnitude of the charge, and r is the distance between the point and the charges.

2. How does the electric potential change as the distance from the point charges increases?

The electric potential decreases as the distance from the point charges increases. This is because as the distance increases, the electric field strength decreases and therefore, the potential energy decreases.

3. Can the electric potential be negative between two identical + 8.5mu C point charges?

Yes, the electric potential can be negative between two identical + 8.5mu C point charges. This occurs when the point is closer to one of the charges than the other, causing the electric potential to be more negative due to the stronger electric field.

4. How does the magnitude of the point charges affect the electric potential?

The magnitude of the point charges has a direct effect on the electric potential. As the magnitude of the charges increases, the electric potential also increases. This is because the electric field strength increases, resulting in a higher potential energy at a given point.

5. Can the electric potential be zero between two identical + 8.5mu C point charges?

Yes, the electric potential can be zero between two identical + 8.5mu C point charges. This occurs when the point is equidistant from both charges, resulting in equal and opposite electric fields that cancel each other out, resulting in a net electric potential of zero.

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