What Was the Initial Charge on Sphere A in the Conducting Spheres Problem?

[a85liz]

Homework Statement

Figure 21-36 shows four identical conducting spheres that are actually well separated from one another.

Figure 21-36
Sphere W (with an initial charge of zero) is touched to sphere A and then they are separated. Next, sphere W is touched to sphere B (with an initial charge of -29e) and then they are separated. Finally, sphere W is touched to sphere C (with an initial charge of +43e), and then they are separated. The final charge on sphere W is +13e. What was the initial charge on sphere A?

Homework Equations

don't know

The Attempt at a Solution

I tried working backwards from the known charge and using the formula Q=.5(q1-q2). I am sure I am missing something simple, but I can not figure this problem out.

 

[LowlyPion]

Homework Statement

Figure 21-36 shows four identical conducting spheres that are actually well separated from one another.

Sphere W (with an initial charge of zero) is touched to sphere A and then they are separated. Next, sphere W is touched to sphere B (with an initial charge of -29e) and then they are separated. Finally, sphere W is touched to sphere C (with an initial charge of +43e), and then they are separated. The final charge on sphere W is +13e. What was the initial charge on sphere A?

I tried working backwards from the known charge and using the formula Q=.5(q1-q2). I am sure I am missing something simple, but I can not figure this problem out.

Welcome to PF.

Maybe try working forwards? After touching A you know W carries 1/2A.

You know what B carries, so after touching B you have 1/2(1/2A + (-29))

So develop the formula for after touching C and knowing the final value, solve for A.

 

[nlightner]

As a scientist, it is important to first identify and understand the problem before attempting to solve it. In this case, we are given information about four conducting spheres and their interactions with each other. The final goal is to determine the initial charge on sphere A.

To begin, it is important to note that when two conducting objects are brought into contact, their charges will redistribute until they reach an equilibrium state. This means that the total charge before and after the contact remains the same. Using this principle, we can set up an equation to represent the initial and final charges on sphere W.

Let QW be the initial charge on sphere W, QA be the initial charge on sphere A, QB be the initial charge on sphere B, and QC be the initial charge on sphere C. We know that the final charge on sphere W is +13e, so we can write:

QW + QA + QB + QC = +13e

Next, we can use the information given about the interactions between the spheres to set up equations for each contact. When sphere W is touched to sphere A, their charges will redistribute and we can use the formula Q=.5(q1-q2) to represent this. This gives us the equation:

QW = 0.5(QA - 0)

Similarly, when sphere W is touched to sphere B, we have:

QW = 0.5(QB - 29e)

And when sphere W is touched to sphere C, we have:

QW = 0.5(QC + 43e)

We now have a system of equations representing the initial and final charges on sphere W. We can solve this system of equations to find the initial charges on each sphere. Substituting the first equation into the second and third equations, we get:

0.5(QA - 0) = 0.5(QB - 29e)
0.5(QA - 0) = 0.5(QC + 43e)

Simplifying these equations, we get:

QA = QB - 29e
QA = QC + 43e

Now, we can set these two equations equal to each other and solve for QB:

QB - 29e = QC + 43e
QB = QC + 72e

Finally, substituting this value for QB into our original equation for the final charge on sphere W, we get:

QW + QA + (QC +