Isn't $\varepsilon_0\int_{\text{all space}}\vec{E}_1\cdot\vec{E}_2 \,{\rm d}v$ just the potential energy?

Question

I have two metallic spheres each with a charge of $q_1$ and $q_2$ respectively. What is the value of $$\varepsilon_0\int_{\text{all space}} \vec{E}_1\cdot\vec{E}_2 \,{\rm d}v$$

where $\vec{E}_1$ and $\vec{E}_2$ are the electric fields due to the two spheres and $\varepsilon_0$ is the permittivity of free space. The radius of the two spheres are $a$ and $b$ and the distance between their centers is $R$ with $R>>>a,b$

(As far as I know), $\frac{1}{2}\varepsilon_0\vec{E}_1\cdot \vec{E}_2$ would be the electric field energy density at any point, so this integral should be double of the net electrostatic interaction potential energy (as opposed to self-potential energy) between the two bodies. So I thought that it would be $$\frac{2q_1q_2}{4\pi\varepsilon_0R},$$

but according to the answer key in my test it is actually

$$\frac{q_1q_2}{4\pi\varepsilon_0R}.$$

How? Where am I wrong?

Answer 1

The total electric field in all of space is $\vec{E} = \vec{E}_1 + \vec{E}_2$. The electric field energy density at a point is $$ \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 E_1^2 + \frac{1}{2} \epsilon_0 E_2^2 + \epsilon_0 \vec{E}_1 \cdot \vec{E}_2 $$ The first two contributions are the self-energies of the electric fields, whereas the third contribution is the contribution due to the interaction. Thus, $\epsilon_0 \vec{E}_1 \cdot \vec{E}_2$ is the electric field energy density due to the interaction (not $\frac{1}{2} \epsilon_0 \vec{E}_1 \cdot \vec{E}_2$). Integrating this over all space gives us the potential energy of the interaction, which is $$ V = \frac{q_1 q_2}{4\pi \epsilon_0 R} $$