According to Einstein energy gravitates, so both the electrostatic field energy as well as the gravitational field energy associated with charged particles must gravitate. Their effect on the gravitational acceleration experienced by charged particles, $g_E$, is given by Equation (7) in [4] :

$g_E=\frac{\left(1-\frac{e^2}{c^{2}mr}-\frac{Gm}{c^{2}r}-\frac{GM}{c^2}-\frac{3{\stackrel{\dot{}}{r}}^2}{2c^2}\right)g_N}{1+\frac{3GM}{c^{2}r}+\frac{3{\stackrel{\dot{}}{r}}^2}{2c^2}}$ (1)

$G$ is the gravitational constant, $e$ the charge of the particle, $c$ is the speed of light, $m$ the mass of the particle experiencing gravitational acceleration, $M\gg m$, $r$ is the distance to the center of $M$ and and $\stackrel{\dot{}}{r}$ is the radial velocity of $m$, which we set equal to zero because we are only interested in the gravitational effect of electrostatic field energy.

According to the above equation for an electron of mass, $m_e$, on the surface of the earth, the effect of its electrostatic field energy on the gravitational acceleration is:

$g_e=-\frac{e^2}{m_e{c}^{2}r}{g}_N=-4.4\times {10}^{-22}{g}_N=-\eta g_N$ (2)

which is the classical electron radius divided by the radius of the earth, $r$,

multiplied by $g_N=-\frac{GM}{r^2}$, the Newtonian acceleration of gravity. $e=1.6\times {10}^{-19}$

is the charge of an electron. It is important to note that $g_e>0$ because $g_N<0$, which means that electrostatic field energy gravitates repulsively. For convenience we define: $\eta =4.4\times {10}^{-22}$.

If an electrically charged body of mass, $m$, contains $n$ electrons, it follows from the above equations that the contribution of the electrostatic field energy, $g_E$, to the gravitational acceleration is:

$g_E=\frac{n^2}{m}{g}_e$ (3)

Our task is the determine how many electrons in the hollow space of the insulator are required for the gravitational effect of electrostatic field energy to be detectable in a torsion balance instrument. To accomplish this task we employ Equation (3).

$g_E$ in Equation (3) is the contribution of the electrostatic field energy to the gravitational acceleration of the test body that contains electrons in the hollow space of the insulator. n is the number of electrons in the hollow space. m is the mass of this test body (mass of conducting outer shell, $m_{cs}$ + mass of insulator shell, $m_{is}$ + mass of electrons added, $n$). m is in units of electron mass. So we must convert $m_{cs}$ and $m_{is}$ in kilograms to electron mass units by dividing them by the electron mass in kilograms, $m_e=9.1\times {10}^{-31}$.

$m=\frac{m_{cs}+m_{is}}{m_e}+n$ (4)

The test body, which does not contain additional electrons, experiences the Newtonian gravitational acceleration, $g_N$. The torsion balance instrument mea-sures the difference in gravitational acceleration between these two test bodies. This difference is $g_E$. There is however a limit to the precision, $\delta$, any experiment can achieve. By $\delta$ we mean the smallest difference in the gravitational acceleration between the two test bodies that can be detected. Clearly, $g_E\ge \delta$.

For simplicity of calculation we assume both conductors and insulators are perfect spherical shells of uniform density, located in a vacuum meaning we do not have to worry about breakdown in air. In this section we calculate the range of radii and masses of the test bodies that will allow us to detect the gravitational effect of electrostatic field energy in a torsion balance instrument as a function of the number of electrons, $n$, in the hollow space of the insulator. As depicted in Figure 1 our test bodies have four sections: (1) hollow space inside the insulator shell (2) insulator shell (3) gap between the insulator shell and the conducting shell (4) conducting outer shell. We discuss the physical properties of each section separately.

Inserting Equation (2) into Equation (3) we obtain:

$g_E=-\eta \frac{n^2}{m}{g}_N$ (5)

Because $\eta =4.4\times {10}^{-22}$ is a very small number it is clear from the above equation that n must be large in order for $g_E$ to be detectable. Consequently, we need to maximize the number of electrons in this space to insure the detectability of the gravitational acceleration caused by the electrostatic field energy of the electrons.

The pertinent quantities of the hollow space are: $n$, number of electrons in this hollow space and $R$, radius of the hollow space, which is also the inner radius of the insulator. The maximum number of electrons is limited by the electric field at the inner surface of the insulator shell, which must not exceed the dielectric breakdown, $E_{\max }$, of the insulator shell. From the theory of electrostatics the maximum allowable electric field, $E$, at the surface of the insulator is:

$E=\frac{1}{4\pi {ϵ}_0}\frac{Q}{R^2}=\frac{\mu n}{R^2}=E_{\max }$ (6)

where the charge, $Q$, is: $Q=ne$ and $\mu =\frac{e}{4\pi {ϵ}_0}$.

$n$ is the primary factor in determining the effect of the gravitational acceleration, $g_E$, caused by the electrostatic field energy of the electrons in the hollow space of the insulator. So we rearrange the above equation to obtain an expression for $n$.

$n=\frac{1}{\mu }{R}^2{E}_{\max }$ (7)

As the previous section makes clear the number of electrons in the hollow space is large. Consequently, there will be strong electric fields at the boundary of the hollow space. We therefore suggest that the space that contains the electron gas be bounded by an insulator. If there is no insulator then field emission, discharge, electronic tunneling into the conductor or even runaway charge loss could occur. An insulator avoids these possibilities.

Equation (7) shows that $n\propto E_{\max }$. We need $n$ to be large. Therefore, we require that the dielectric strength, $E_{\max }$, be as large as possible. Among the common insulators with the highest dielectric strengths (Teflon, quartz, fused silica, diamond) quartz has the highest value: $E_{\max }={10}^9\text{V}/\text{m}$. So baring other considerations, we suggest employing quartz in our proposed experiment to determine the gravitational effect of electrostatic field energy.

We need to compute the properties of the insulator shell (mass, radius and volume) as a function of the number of electrons in the hollow space it encloses. The mass of the insulator shell is determined by the density, ${\rho }_q=2650\text{kg}/{\text{m}}^{\text{3}}$, inner radius and width of the shell. The inner radius is the radius of the hollow space enclosed by the insulator, R, which we now denote as $R_q$ because we have determined that the insulator should be made out of quartz. It depends upon the number of electrons in the hollow space. Rearranging Equation (7) gives of an equation for $R_q$.

$R_q=\sqrt{\frac{\mu n}{E_{\max }}}$ (8)

The expressions for the volume and mass of the insulator depend upon the width of the insulator. What should the width of the insulator, $\text{Δ}{R}_q$, be? At the outer edge of the insulator is a gap. Therefore, the electric field at the outer edge, $R_q+\text{Δ}{R}_q$, of the insulator should be below the vacuum breakdown, $E_{\text{vac}}$. Gauss’s Law states:

$E_{\text{vac}}=\frac{\mu n}{{\left(R_q+\text{Δ}{R}_q\right)}^2}$ (9)

Solving this equation for $\text{Δ}{R}_q$ yields:

$\text{Δ}{R}_q=\sqrt{\frac{\mu n}{E_{\text{vac}}}}-R_q$ (10)

Inserting $R_q$ from Equation (8) into the above equation yields $\text{Δ}{R}_q=0$. This result means that $\text{Δ}{R}_q$ is small. Experience shows that it can be just a few millimeters.

The equations for the volume and mass of the quartz insulator are:

$V_q=\frac{4}{3}\pi \left({\left(R_q+\text{Δ}{R}_q\right)}^3-R_q^3\right)=4\pi \left(\frac{1}{3}\text{Δ}{R}_q^3+\sqrt{\frac{\mu n}{E_{\text{max}}}}\text{Δ}{R}_q^2+\frac{\mu n}{E_{\text{max}}}\text{Δ}{R}_q\right)$ (11)

$m_q={\rho }_q{V}_q=4\pi {\rho }_q\left(\frac{1}{3}\text{Δ}{R}_q^3+\sqrt{\frac{\mu n}{E_{\text{max}}}}\text{Δ}{R}_q^2+\frac{\mu n}{E_{\text{max}}}\text{Δ}{R}_q\right)$ (12)

We suggest the presence of a vacuum gap between the insulator and the conducting shell for the following reasons. Even though the insulator is not charged, the charges in the electron gas create an electric field that would extend through the insulator to the conductor interface, if there were no gap. At that interface, field intensification can still occur due to: field focusing, local surface imperfections and permittivity boundary effects. The existence of the gap will allow the conducting shell to induce exactly the right charge on its inner surface to cancel the electric field from the electrons in the hollow space of the insulator. Consequently, the outer surface will remain field-free, which is the primary reason for having a conducting shell. We now turn to answering the question: What is an appropriate gap width?

The width of the gap, $g_w$, must be large enough to prevent dielectric breakdown of the vacuum. Specifically, the gap between the quartz shell and the conductor shell must be large enough to avoid field emission, corona discharge, or tunneling effects. We can estimate the minimum gap width, $g_w$, by employing the equation:

$g_w=\sqrt{\frac{\mu n}{E_{\text{Al}}}}-\sqrt{\frac{\mu n}{E_{\text{vac}}}}$ (13)

where $E_{\text{Al}}$ is breakdown threshold of aluminum. In the next section we discuss the reason we suggest aluminum.

In order to eliminate the prodigious electric fields, $E\approx {10}^9\text{V}/\text{m}$, outside the insulator, which would not allow the high precision required, we suggest encompassing the insulator and gap in a conducting shell. Remembering that $g_E\propto m^{-1}$, it is important to choose material for the conducting shell so that m is as small as possible. We therefore suggest aluminum, which has the lowest density ${\rho }_{Al}=2700\text{kg}/{\text{m}}^{\text{3}}$, among the common conductors. We now turn to the questions: What should the width, volume and mass of the aluminum shell be?

The aluminum shell has two radii that are needed to compute the volume and mass of the aluminum shell. They are: $R_{\text{Ali}}$ and $R_{\text{Alo}}$, which are the inner and outer radius respectively. We have:

$R_{\text{Ali}}=R_q+\text{Δ}{R}_q+g_w=\sqrt{\frac{\mu n}{E_{\text{max}}}}+\sqrt{\frac{\mu n}{E_{\text{Al}}}}-\sqrt{\frac{\mu n}{E_{\text{vac}}}}+\text{Δ}{R}_q$ (14)

For quartz: $E_{\max }={10}^9\text{V}/\text{m}$ and $E_{\text{vac}}={10}^9\text{V}/\text{m}$. So the above equation reduces to:

$R_{\text{Ali}}=\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q$ (15)

for the outer radius of the aluminum shell:

$R_{\text{Alo}}=R_{\text{Ali}}+\text{Δ}{R}_{\text{Al}}=\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q+\text{Δ}{R}_{\text{Al}}$ (16)

where $\text{Δ}{R}_{\text{Al}}$ is the thickness of the aluminum shell. Like the thickness of the insulator $\text{Δ}{R}_{\text{Al}}$ is just a few millimeters.

The mass is obtained from the volume, $V_{\text{Al}}$, and density, ${\rho }_{\text{Al}}$. That is:

$V_{\text{Al}}=\frac{4}{3}\pi \left(R_{\text{Alo}}^3-R_{\text{Ali}}^3\right)=\frac{4}{3}\pi \left({\left(\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q+\text{Δ}{R}_{\text{Al}}\right)}^3-{\left(\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q\right)}^3\right)$ (17)

$m_{\text{Al}}={\rho }_{\text{Al}}{V}_{\text{Al}}=\frac{4\pi {\rho }_{\text{Al}}}{3}\left({\left(\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q+\text{Δ}{R}_{\text{Al}}\right)}^3-{\left(\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q\right)}^3\right)$ (18)

We recall that $g_E\ge \delta$. $g_E$ is the gravitational acceleration generated by the electrostatic field energies of the electrons located in the hollow space of the quartz insulator. The precision, $\delta$, is the smallest difference in the gravitational acceleration between the two test bodies that can be detected.

First, we calculate the $g_E$ required for the gravitational effect of the electrostatic field energy of the electrons in the hollow space of the quartz insulator to be detectable. It is a function of the number of electrons, $n$, that are in this hollow space. To accomplish this task we insert Equation (4) into Equation (3), whereby $m_{is}=m_q$ and $m_{cs}=m_{\text{Al}}$ because we have determined that the insulator should be quartz and the conductor aluminum.

$g_E=-\frac{\eta n^2}{\frac{m_q+m_{\text{Al}}}{m_e}+n}{g}_N$ (19)

Insertion of expressions for $m_q$ from Equation (12) and for $m_{\text{Al}}$ from Equation (18) into the above equation gives us an explicit connection between the gravitational acceleration, $g_E$, caused by the electrostatic field energies of the hollow space electrons and $n$, the number of electrons in the hollow space. The calculated $g_E$ is also the precision, $\delta$, required for the gravitational effect of the electrostatic field energies of n electrons to be detectable.

$g_E=-\frac{\eta n^2}{\frac{4\pi {\rho }_q\left(\frac{1}{3}\text{Δ}{R}_q^3+\sqrt{\frac{\mu n}{E_{\text{max}}}}\text{Δ}{R}_q^2+\frac{\mu n}{E_{\text{max}}}\text{Δ}{R}_q\right)+\frac{4\pi {\rho }_{\text{Al}}}{3}\left({\left(\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q+\text{Δ}{R}_{\text{Al}}\right)}^3-{\left(\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q\right)}^3\right)}{m_e}+n}{g}_N$ (20)

Figure 2 is a plot of ${\text{Log}}_{10}\left(g_E\right)$ or ${\text{Log}}_{10}\left(\delta \right)$ vs. ${\text{Log}}_{10}\left(n\right)$ calculated from the above equation, where $g_E$ and $\delta$ are in m/s². In order to construct it, we assumed that $\Delta R_{\text{Al}}=\Delta R_q=0.005\text{m}$.

Figure 3 shows the mass, ${\text{Log}}_{10}\left(m_q\right)$, in kilograms of the quartz shell (Equation (12)) as a function of ${\text{Log}}_{10}\left(n\right)$.

Figure 4 shows the mass, ${\text{Log}}_{10}\left(m_{\text{Al}}\right)$, in kilograms of the aluminum shell (Equation (18)) as a function of ${\text{Log}}_{10}\left(n\right)$.

Figure 5 depicts the inner radius, ${\text{Log}}_{10}\left(R_q\right)$, in meters of the quartz shell in meters (Equation (8)) as a function of ${\text{Log}}_{10}\left(n\right)$. $R_q$ is also the radius of the hollow space that contains the electron gas.

Figure 6 shows the outer radius, ${\text{Log}}_{10}\left(R_{\text{Alo}}\right)$, in meters of the aluminum shell (Equation (16)) as a function of ${\text{Log}}_{10}\left(n\right)$.

Finally, Figure 7 depicts the width of the gap, $g_w$, in meters (Equation (13)) as a function of ${\text{Log}}_{10}\left(n\right)$.

In this section we derive what the precision, $\delta$, must at least be in order to detect the effect of the gravitational acceleration caused by electrostatic field energy. We also derive the number of electrons, $n$, in the electron gas and the properties of the quartz and aluminum bodies (mass and radius) that are required in order to insure that the proposed experiment is feasible.

First, we explore, if our test bodies could fit in the [11] experiment, which is currently operating. This experiment achieves: $\delta =8.8\times {10}^{-15}\text{m}/{\text{s}}^{\text{2}}$ or ${\text{Log}}_{10}\left(\delta \right)=-14.1$. Remembering that $g_E\ge \delta$ it follows that $g_E=\delta$ is the minimum value $g_E$ must have in order for the gravitational acceleration caused by electrostatic field energy to be detectable. Because all physical properties of our test bodies are a function of $n$, the next step is to solve Equation (20) for n in order to derive the physical characteristics of our test bodies. However, it is a transcendental equation and is therefore, not algebraically solvable. From Figure 2 or more directly from Equation (20) we see that $g_E=\delta =8.8\times {10}^{-15}$ corresponds to: $n={10}^{21.7}=5\times {10}^{21}$ electrons. Starting with this value we employ our equations to derive the other properties of the system.

$R_q=84.9\text{m}$ (Equation (8)), $m_q=1.2\times {10}^6\text{kg}$ (Equation (12)), $m_{\text{Al}}=1.2\times {10}^7\text{kg}$ (Equation (18)), $R_{\text{Alo}}=268.5\text{m}$ (Equation (16)). Clearly, such high values means the current experiment can not determine the gravitational acceleration due to the electrostatic field energies of electrons in the hollow space of such a test body. These values also tell us, what is required to detect the effect if the precision is $\delta =8.8\times {10}^{-15}$.

It appears that the experiment is only feasible, if it is possible to decrease by orders of magnitude the value of $\delta$, which is the smallest difference in the gravitational acceleration between the two test bodies that can be detected. In order to see about what the precision must be for laboratory masses, we assume $m=20\text{g}$ in Equation (3). The expression for $m$ is:

$m=m_q+m_{\text{Al}}+n{m}_e$ (21)

Inserting Equations (12) and (18) into the above equation leads to $m$ in kilograms:

$\begin{array}{l}m=4\pi {\rho }_q\left(\frac{1}{3}\text{Δ}{R}_q^3+\sqrt{\frac{\mu n}{E_{\text{max}}}}\text{Δ}{R}_q^2+\frac{\mu n}{E_{\text{max}}}\text{Δ}{R}_q\right)\\ +\frac{4\pi {\rho }_{\text{Al}}}{3}\left({\left(\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q+\text{Δ}{R}_{\text{Al}}\right)}^3-{\left(\sqrt{\frac{\mu n}{E_{\text{Al}}}}+\text{Δ}{R}_q\right)}^3\right)+n{m}_e\end{array}$ (22)

The numerical solution for $m=20\text{g}$ is: $n=4.8\times {10}^{11}$. This value of n leads to:

$R_q=0.84\text{mm}$ (Equation (8)), $m_q=2.2\text{g}$ (Equation (12)), $m_{\text{Al}}=17.8\text{g}$ (Equation (18)), $R_{\text{Alo}}=12.6\text{m}$ (Equation (16)). Finally, we calculate the crucial quantity, $g_E$ by inserting the value for $n=4.8\times {10}^{11}$ in Equation (20). We obtain: $g_E=4.6\times {10}^{-26}$. So the precision must be no larger than $\delta =g_E=4.6\times {10}^{-26}$ to detect the effect of the gravitational acceleration caused by the electrostatic field energy of electron gas containing $n=4.8\times {10}^{11}$ electrons.

This value of $\delta =g_E=4.6\times {10}^{-26}$ may not be achievable. In order to detect this effect with a $\delta$ value larger than this value one must increase all of the physical quantities associated with the test bodies. Specifically, Figure 2 shows that if we increase the value of $g_E$ and correspondingly $\delta$ then the values of all the properties of the test bodies must increase too as Figures 3-6 make graphically clear. We conclude it may not be possible to perform this experiment with laboratory size test bodies.

We have shown how critically important the precision, $\delta$, is for the success of our experiment. To emphasize this point in Table 1 , we list the physical properties of the test bodies as a function of $\delta$.

The success of our proposal depends upon over coming practical challenges. (1) dielectric breakdown of the quartz insulator: This can be mitigated with the use of ultra-pure quartz. (2) charge instabilities: This may be mitigated by employing hybrid Penning geometry. (3) vacuum issues: mitigation through cryopumping, ion getters, or a multi-stage vacuum chamber. (4) Emission: This can be mitigated through the use of ultra-polished quartz. Finally, we note the major source of false effects in a torsion balance experiment comes from gravity gradients, which are thoroughly discussed in [12] .