# electric potenial energy problem

## electric potenial energy problem

four particles of mass m and charge Q are placed at the vertices of the square of sideL. if they are released from their vertices calculate their speeds when the distance from the center of square doubles

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

## Re: electric potenial energy problem

It is very easy to accidentally count the potential energy twice. You must remember that the potential energy is a property of the system as a whole, not of the individual charges. For example, here is a WRONG way to think about it:

Each charge sees the potential from the other 3 charges, which is V = kq/L + kq/(L*sqrt(2)) + kq/L = kq/L*(2 + 1/sqrt(2)). Therefore it has a potential energy qV = kq^2/L*(2 + 1/sqrt(2)). Since there are 4 charges, the total potential energy is 4kq^2/L*(2+ 1/sqrt(2)). Later, when the distance is doubled to 2L, the potential energy is 4kq^2/(2L)*(2+ 1/sqrt(2)). Subtracting these gives the change in potential energy, which will be the kinetic energy, so 4*(1/2*mv^2) = 4kq^2/L*(2+ 1/sqrt(2)) - 4kq^2/(2L)*(2+ 1/sqrt(2)) = 4kq^2/(2L)*(2+ 1/sqrt(2)). Therefore, v = q*sqrt(k/L*(2+1/sqrt(2))).

What is wrong with this method? It was wrong when I said "it has a potential energy" because that is not true. The charge does not have a potential energy that belongs to it individually. Potential energy is like a joint checking account at a bank. If two people A and B have a bank account together, and the account has $100 in it, then A can spend $100 OR B can spend $100, but then cannot BOTH spend $100. They each have ACCESS to the amount of money in the bank, but only if the other one does not spend it.

What this means for energy is that the formula kq^2/L*(2 + 1/sqrt(2)) tells you the amount of kinetic energy that one charge can take from the system if NONE of the other charges takes any. Of course, that is not what is happening in this problem, because all 4 charges are taking some energy. Since all the charges move away from each other, the electric force between them decreases more rapidly than in the case where only one charge moves away, and so the force does less work on each charge, giving it less energy.

This may sound very complicated to deal with, but there is a simple solution. Consider a simpler system with just 2 charges in it. Then the potential is kq/L, and each charge has access to a potential energy kq^2/L. Suppose you hold one charge in place and let the other go. When it is far away, it will get a kinetic energy 1/2*mv^2 = kq^2/L. Now what happens when you release the other charge? Nothing happens, because the first charge is already far away, so there is no force on the second charge. Therefore the second charge gets no kinetic energy. This means that the first charge ran away with ALL the energy. Of course, if you had held the second charge and released the first charge instead, then that one would have run away with all the energy.

Now you can easily see what is wrong with adding the energies of the individual charges: you are seeing the SAME energy twice, once from the point of view of the first charge and once from the point of view of the second charge. There is really only one energy here, just as there is really only $100 in the bank account. If you label the charges q1 and q2, it is easy to see what is happening: q1 makes a potential V1 = kq1/L, and this potential gives q2 a potential energy q2V1. Similarly, q2 makes a potential V2 and q1 has a potential energy q1V2. It looks very reasonable to add them, q1V2 + q2V1, but if you look at what each formula actually says, q1V2 = kq1q2/L = kq2q1/L = q2V1, so this is just the same energy by two different names. When you add the energies like this, you are always seeing the same energy exactly twice. This is because each charge appears in the formula once as a SOURCE of V and once as a RECEIVER of V.

What happens when you have several charges? There are two ways to calculate the energy. One way is to look at all the PAIRS of charges and add their energies. In the case with 4 charges, the pairs would be q1q2, q1q3, q1q4, q2q3, q2q4, q3q4. Suppose q1 and q3 are on opposite corners of the square. Then the energy would be

kq1q2/L + kq1q3/(Lsqrt(2)) + kq1q4/L + kq2q3/L + kq2q4/(Lsqrt(2)) + kq3q4/L = 4kq^2/L + 2kq^2/(Lsqrt(2)).

Notice that this is exactly HALF the answer that I got by the wrong method earlier.

This leads to the second way to calculate the energy. Look again at the wrong method. If you label the charges and write out in detail all the individual calculations, it looks like this:

Potential energy of q1 = U1 = q1(kq2/L + kq3/(Lsqrt(2)) + kq4/L)

Potential energy of q2 = U2 = q2(kq1/L + kq3/L + kq4/(Lsqrt(2)))

Potential energy of q3 = U3 = q3(kq1/(Lsqrt(2)) + kq2/L + kq4/L)

Potential energy of q1 = U1 = q4(kq1/L + kq2/(Lsqrt(2)) + kq3/L)

Now if you add these together (as in the wrong method) and factor out the k/L, you get

U_total =

k/L*(q1q2 + q1q3/sqrt(2) + q1q4 + q2q1 + q2q3 + q2q4/sqrt(2) + q3q1/sqrt(2) + q3q2 + q3q4 + q4q1 + q4q2/sqrt(2) + q4q3) =

k/L((q1q2+q2q1) + (q1q3+q3q1)/sqrt(2) + (q1q4+q4q1) + (q2q3+q3q2) + (q2q4+q42)/sqrt(2) + (q3q4+q4q3)).

Do you see what happened here? Each pair of terms in parenthesis is a pair of charges showing the same mistake that happened in the simple two-charge system. For instance q1q2+q2q1 is showing the same energy counted twice, the first time when q1 is receiving the potential from q2 and the second time when q2 is receiving the potential from q1. So you see, in the wrong method, all the energy is counted exactly twice.

Therefore the second way to calculate the energy is very easy: just use the wrong method and divide by 2. This is GUARANTEED to get the correct energy, no matter how many charges there are, because they will always be grouped into pairs, and each pair will count the same energy twice.

I am sorry that this explanation is so LONG, but I wanted to show you exactly WHY you need to divide by 2 to get the book's answer. In my experience, many textbooks do not do a thorough enough job explaining how to count the total potential energy of the system.

Hope this helps!

--Pinnaka

Note: this problem has been answered by our expert pinnaka

Each charge sees the potential from the other 3 charges, which is V = kq/L + kq/(L*sqrt(2)) + kq/L = kq/L*(2 + 1/sqrt(2)). Therefore it has a potential energy qV = kq^2/L*(2 + 1/sqrt(2)). Since there are 4 charges, the total potential energy is 4kq^2/L*(2+ 1/sqrt(2)). Later, when the distance is doubled to 2L, the potential energy is 4kq^2/(2L)*(2+ 1/sqrt(2)). Subtracting these gives the change in potential energy, which will be the kinetic energy, so 4*(1/2*mv^2) = 4kq^2/L*(2+ 1/sqrt(2)) - 4kq^2/(2L)*(2+ 1/sqrt(2)) = 4kq^2/(2L)*(2+ 1/sqrt(2)). Therefore, v = q*sqrt(k/L*(2+1/sqrt(2))).

What is wrong with this method? It was wrong when I said "it has a potential energy" because that is not true. The charge does not have a potential energy that belongs to it individually. Potential energy is like a joint checking account at a bank. If two people A and B have a bank account together, and the account has $100 in it, then A can spend $100 OR B can spend $100, but then cannot BOTH spend $100. They each have ACCESS to the amount of money in the bank, but only if the other one does not spend it.

What this means for energy is that the formula kq^2/L*(2 + 1/sqrt(2)) tells you the amount of kinetic energy that one charge can take from the system if NONE of the other charges takes any. Of course, that is not what is happening in this problem, because all 4 charges are taking some energy. Since all the charges move away from each other, the electric force between them decreases more rapidly than in the case where only one charge moves away, and so the force does less work on each charge, giving it less energy.

This may sound very complicated to deal with, but there is a simple solution. Consider a simpler system with just 2 charges in it. Then the potential is kq/L, and each charge has access to a potential energy kq^2/L. Suppose you hold one charge in place and let the other go. When it is far away, it will get a kinetic energy 1/2*mv^2 = kq^2/L. Now what happens when you release the other charge? Nothing happens, because the first charge is already far away, so there is no force on the second charge. Therefore the second charge gets no kinetic energy. This means that the first charge ran away with ALL the energy. Of course, if you had held the second charge and released the first charge instead, then that one would have run away with all the energy.

Now you can easily see what is wrong with adding the energies of the individual charges: you are seeing the SAME energy twice, once from the point of view of the first charge and once from the point of view of the second charge. There is really only one energy here, just as there is really only $100 in the bank account. If you label the charges q1 and q2, it is easy to see what is happening: q1 makes a potential V1 = kq1/L, and this potential gives q2 a potential energy q2V1. Similarly, q2 makes a potential V2 and q1 has a potential energy q1V2. It looks very reasonable to add them, q1V2 + q2V1, but if you look at what each formula actually says, q1V2 = kq1q2/L = kq2q1/L = q2V1, so this is just the same energy by two different names. When you add the energies like this, you are always seeing the same energy exactly twice. This is because each charge appears in the formula once as a SOURCE of V and once as a RECEIVER of V.

What happens when you have several charges? There are two ways to calculate the energy. One way is to look at all the PAIRS of charges and add their energies. In the case with 4 charges, the pairs would be q1q2, q1q3, q1q4, q2q3, q2q4, q3q4. Suppose q1 and q3 are on opposite corners of the square. Then the energy would be

kq1q2/L + kq1q3/(Lsqrt(2)) + kq1q4/L + kq2q3/L + kq2q4/(Lsqrt(2)) + kq3q4/L = 4kq^2/L + 2kq^2/(Lsqrt(2)).

Notice that this is exactly HALF the answer that I got by the wrong method earlier.

This leads to the second way to calculate the energy. Look again at the wrong method. If you label the charges and write out in detail all the individual calculations, it looks like this:

Potential energy of q1 = U1 = q1(kq2/L + kq3/(Lsqrt(2)) + kq4/L)

Potential energy of q2 = U2 = q2(kq1/L + kq3/L + kq4/(Lsqrt(2)))

Potential energy of q3 = U3 = q3(kq1/(Lsqrt(2)) + kq2/L + kq4/L)

Potential energy of q1 = U1 = q4(kq1/L + kq2/(Lsqrt(2)) + kq3/L)

Now if you add these together (as in the wrong method) and factor out the k/L, you get

U_total =

k/L*(q1q2 + q1q3/sqrt(2) + q1q4 + q2q1 + q2q3 + q2q4/sqrt(2) + q3q1/sqrt(2) + q3q2 + q3q4 + q4q1 + q4q2/sqrt(2) + q4q3) =

k/L((q1q2+q2q1) + (q1q3+q3q1)/sqrt(2) + (q1q4+q4q1) + (q2q3+q3q2) + (q2q4+q42)/sqrt(2) + (q3q4+q4q3)).

Do you see what happened here? Each pair of terms in parenthesis is a pair of charges showing the same mistake that happened in the simple two-charge system. For instance q1q2+q2q1 is showing the same energy counted twice, the first time when q1 is receiving the potential from q2 and the second time when q2 is receiving the potential from q1. So you see, in the wrong method, all the energy is counted exactly twice.

Therefore the second way to calculate the energy is very easy: just use the wrong method and divide by 2. This is GUARANTEED to get the correct energy, no matter how many charges there are, because they will always be grouped into pairs, and each pair will count the same energy twice.

I am sorry that this explanation is so LONG, but I wanted to show you exactly WHY you need to divide by 2 to get the book's answer. In my experience, many textbooks do not do a thorough enough job explaining how to count the total potential energy of the system.

Hope this helps!

--Pinnaka

Note: this problem has been answered by our expert pinnaka

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

## Re: electric potenial energy problem

We have to calculate the potential energy of the system by analyzing how work one has to do assemble charges from infinity to that location. If take one charge to be fixed at a certain location, we have to calculate the work done to assemble the other three charges to find the potential energy of the system.

Proceeding with the given terms in the problem we get the initial energy of the system to be as follows:

Uinitial = 4keQ^2/l +2 keQ^2/2^(1/2) and the final potential energy of the system is as follows:

Ufinal = 2keQ^2/l + keQ^2/2((2)^(1/2))

Solving the following equation

Change in the potential energy of the system = increase in the kinetic energy of the system

I get the velocity of the individual particle to be

v = sqrt { keQ^2/ml{1+(1/8^(1/2))}}}

Proceeding with the given terms in the problem we get the initial energy of the system to be as follows:

Uinitial = 4keQ^2/l +2 keQ^2/2^(1/2) and the final potential energy of the system is as follows:

Ufinal = 2keQ^2/l + keQ^2/2((2)^(1/2))

Solving the following equation

Change in the potential energy of the system = increase in the kinetic energy of the system

I get the velocity of the individual particle to be

v = sqrt { keQ^2/ml{1+(1/8^(1/2))}}}

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

Similar topics

» Electricity: a plan to set up two stations and we call on Iran to provide us with energy

» Electric Cooperatives

» Silverfish problem....

» 32 Firms Buy Info-Packs for 4th Energy Round

» Glut cars and energy methods (easy problem to solve)

» Electric Cooperatives

» Silverfish problem....

» 32 Firms Buy Info-Packs for 4th Energy Round

» Glut cars and energy methods (easy problem to solve)

Page

**1**of**1****Permissions in this forum:**

**cannot**reply to topics in this forum