# Question about current and resistance

1. Nov 28, 2011

### pc2-brazil

1. The problem statement, all variables and given/known data

Two isolated conducting spheres, each one with a radius of 14 cm, are charged until they reach the potentials of 240 V and 440 V. They are then connected by a thin wire. Calculate the variation of internal energy of the wire.

2. Relevant equations

3. The attempt at a solution

When the wire is connected to both spheres, charge flows from the 240 V sphere to the 440 V sphere, in order to equal the potential. Since both spheres have the same radius, the final charge in each sphere will be equal. The final potential in each sphere will equal (240 + 440)/2 = 340 V.
I think that the variation of internal energy is related to the fact that the potential difference between the ends of the wire changes from 200 V to 0 V as the charge flows through it.
But I'm not sure how to proceed from here.

Thank you in advance.

2. Nov 28, 2011

### ehild

There is potential difference across the wire. So there is current flowing through it. The flowing current will dissipate energy by warming up the wire. How do you get the power - the energy transferred to the wire in unit time - with the potential difference and the current? How do you get the energy transferred to the wire during the whole process?
The current is the charge q transferred from one capacitor to the other in unit time I=dq/dt. The potential difference can be expressed by this q and the initial charges. How?

ehild

3. Nov 29, 2011

### pc2-brazil

The power is:
$$P=i\Delta V$$
I imagine that the energy transferred to the wire will be the change in electric potential energy as the charges move from one sphere to the other one; since the wire has some resistance, the change in electric potential energy will be converted to heat to warm up the wire. Is this right?
The integral of the power with respect to time will give the work done by the electric field, which will correspond to the energy dissipated in form of heat.
This q and the potential difference appear in the work done by the electric field in the wire to move the charges through the potential difference:
$$W=q\Delta V$$
Now I only have to find the total charge transferred.
The charge, because it is negative, will be transferred from the smaller potential (240 V) sphere to the greater potential sphere (440 V).
The potential of the 240 V sphere increases from V1 = 240 V to V = 340 V, as I said before. So, the potential increases 100 V. The amount of charge that went through the wire will be, then, the final charge of this sphere minus its initial charge:
$$q=\frac{r}{k}V-\frac{r}{k}V_1=\frac{r}{k}(V-V_1)$$
where $k=9\times 10^9$, $r=14\times 10^{-2}$ and V - V1 = 100.
Using the value of q and the value of $\Delta V$ = 200 V, I obtain the correct result:
W = 311.11 nJ.

But I think that the actual amount of work should be smaller than the value I found, because the value of the potential difference accross the wire ($\Delta V$) will not remain constant (200 V) through the whole process. It will continually decrease from 200 V to 0 V as the charge is transferred.
So, I think that the value found is an approximation. Is that correct?

Thank you in advance.

4. Nov 29, 2011

### ehild

Your guess is correct, the potential difference ΔV is changing during the charge transfer so you need this take into account and calculate the work done by integration: W=∫ΔV dq.
The next thing is to use the relationship between the potential difference ΔV and the charge q transferred from the sphere at higher potential to the other one.
At a certain time instant, q charge left the sphere at the higher potential, leaving it with charge Q1-q. The same charge is gained by the other sphere so its charge is Q2+q. What is the potential difference between the spheres then?

ehild

5. Nov 29, 2011

### pc2-brazil

If I call V1 the higher potential and V2 the smaller potential, then:
$$\Delta V = V_1 - V_2$$
In the instant a charge q was transferred from sphere 1 to sphere 2, the value of ΔV is, then:
$$\Delta V = V_1 - V_2 = \frac{k(Q_1 - q)}{r} - \frac{k(Q_2 + q)}{r} = \frac{k}{r}(Q_1 - Q_2 - 2q)$$
The total amount of charge transferred will equal:
$$Q_1 - \frac{Q_1 + Q_2}{2} = \frac{Q_1 - Q_2}{2}$$
So:
$$W = \int \Delta V dq$$
$$W = \frac{k}{r} \int_0^{\frac{Q_1 - Q_2}{2}} (Q_1 - Q_2 - 2q) dq$$
$W = \frac{k}{r} [(Q_1 - Q_2)q - q^2]$ with q going from 0 to $\frac{Q_1 - Q_2}{2}$
$$W=\frac{k}{r}\left [ \frac{(Q_1 - Q_2)^2}{2} - \frac{(Q_1 - Q_2)^2}{4} \right] = \frac{k}{r} \frac{(Q_1 - Q_2)^2}{4}$$
This gives:
$$W=\frac{k}{r}\left [ \frac{(Q_1 - Q_2)^2}{4} \right]$$

Plugging in the values, I get 155.55 nJ, which is approximately half the value I had found without taking into account the variation in the potential difference.
It seems to make sense, but the answer key of the book gives 311 nJ (which is the value I found in the previous reply, neglecting the variation of $\Delta V$, and is approximately twice the value I found here, 155.5 nJ).

Last edited: Nov 29, 2011
6. Nov 29, 2011

### ehild

So either we are wrong or the answer key is. Or it defines "internal energy of wire" in a different way. It is not a a usual concept. It looked straightforward that the "change of internal energy of the wire" was meant as the work done by the electric field during the charge transfer.

Anyway, you did a god job.

It is usually shown during classes that charging a sphere to charge Q and to potential V needs W=QV/2 work. As V=kQ/r, W=V2r/(2k). That is the energy of a charged sphere of radius r.
Similarly, a capacitor has energy E=QΔV/2 = Q2/(2C)=ΔV2C/2.

If you use these formulae available in every textbook, you get that the initial energy of the spheres

$$E(i)=\frac{r}{2k}(V_1^2+V_2^2)$$.

In the final state, V=(V1+V2)/2 for both spheres. So the sum of the final energy of both spheres is

$$E(f)=2 \frac{r}{2k}(\frac{V_1+V_2}{2})^2=\frac{r}{2k} \frac{(V_1+V_2)^2}{2}$$

The decrease of the energy stored by the spheres is

$$E(i)-E(f)= \frac{r}{2k}(V_1^2+V_2^2-\frac{(V_1+V_2)^2}{2})=\frac{r}{k} (\frac {V_1-V_2}{2})^2$$,

and this yields the same W=155 nJ.

ehild

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