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Chapter 6: Problem 5

Show that the work done in bringing two charges \(q_{1}\) and \(q_{2}\), initially far apart, to a separation \(r_{12}\) is \(q_{1} q_{2} / 4 \pi \epsilon_{0} r_{12}\). Write down the corresponding expression for a system of many charges. Show that the energy stored in the charge distribution is $$ V=\frac{1}{2} \sum_{j} q_{j} \phi_{j}\left(\boldsymbol{r}_{j}\right) $$ where \(\phi_{j}\left(\boldsymbol{r}_{j}\right)\) is the potential at \(\boldsymbol{r}_{j}\) due to all the other charges. Why does a factor of \(\frac{1}{2}\) appear here, but not in the corresponding expression for the potential energy in an external potential \(\phi(\boldsymbol{r}) ?\)

Short Answer

Expert verified
Answer: The 1/2 factor appears in the expression for energy stored in a charge distribution because we are considering the interaction between every pair of charges twice when summing up their potential energy. This factor avoids double counting and ensures that the energies of the interacting charges are counted only once. However, this factor does not appear in the potential energy of a charge in an external field, as it involves the energy due to an external field rather than interactions within the system.

Step by step solution

01

Write down the formula of potential energy between two charges

The potential energy of interaction between two charges can be calculated using Coulomb's Law. The formula for potential energy (\(U\)) between two charges \(q_1\) and \(q_2\) separated by a distance \(r_{12}\) is: $$ U = \frac{k q_1 q_2}{r_{12}} $$ Where \(k = \frac{1}{4 \pi \epsilon_{0}}\) is the Coulomb's constant.
02

Calculate the work done by bringing charges closer

Now, to calculate the work (W) done while moving the charges from a far distance to separation \(r_{12}\), we can use the following formula: $$ W = U_{final} - U_{initial} $$ Here, \(U_{final}\) corresponds to the final potential energy between the two charges, and \(U_{initial}\) represents the initial potential energy when the charges are far apart.
03

Evaluate initial and final potential energy

For our problem, the initial potential energy (U_{initial}) will be 0 when the charges are far apart. The final potential energy (U_{final}) is calculated using the formula derived in Step 1: $$ U_{final} = \frac{k q_1 q_2}{r_{12}} $$
04

Calculate the work done

Now, by substituting these initial and final potential energy values in the formula for work done in Step 2, we get: $$ W = \frac{k q_1 q_2}{r_{12}} = \frac{q_1 q_2}{4 \pi \epsilon_{0} r_{12}} $$ This is the required expression for the work done while getting two charges closer to each other with a separation \(r_{12}\).
05

Generalize the expression to multiple charges

To extend this concept to a system with multiple charges, we can calculate the total work done by bringing every pair of charges close together. Now, the total potential energy in a system of many charges will be given by: $$ V = \frac{1}{2} \sum_{i} \sum_{j} \frac{k q_i q_j}{r_{ij}} $$ Here, we sum over every pair of charges i and j, while making sure we don't count the same interaction twice. Therefore, we need to include the factor \(\frac{1}{2}\).
06

Determine the energy stored in the charge distribution

Now, let's rewrite this total potential energy using the expression of potential at a point \(r_j\) due to all other charges: $$ V = \frac{1}{2} \sum_{j} q_j \phi_j(\boldsymbol{r}_j) $$
07

Understand the factor ½ in the energy expression

The factor \(\frac{1}{2}\) appears in the expression for energy when calculating the energy stored in the charge distribution because we are considering the interaction between every pair of charges twice when summing up their potential energy. This factor avoids double counting and makes sure that the energies of the interacting charges are counted only once. In contrast, the factor does not appear in the expression for potential energy in an external potential, as we won't be considering the interaction between charges within the system. Instead, we are focusing on the potential energy of a charge due to an external field.

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Most popular questions from this chapter

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