(a) If the same charge of Q as in part (a) above is given to a spherical conductor of the same radiu

Q: Why is the electric field inside a spherical conductor zero?

In a conductor, charges are free to move. They repel each other and distribute themselves entirely on the outer surface to minimize repulsion, resulting in a net zero electric field inside.

Q: Why do we integrate energy density to find the total energy?

The energy of a charge configuration is stored in the electric field it creates. By integrating the energy density (energy per unit volume) over all space where the field exists, we get the total potential energy of the system.

Q: How is the gravitational self-energy related to the energy needed to disassemble the Earth?

The gravitational self-energy is the work done by gravity to assemble the Earth from infinity, which is a negative value. To disassemble it, we must do an equal amount of positive work against gravity, which is the absolute value of the self-energy.

Q: Why can we replace the electrostatic constants with gravitational ones?

Both Coulomb's law and Newton's law of universal gravitation are inverse-square laws. The mathematical structure of the fields and potentials they create is identical, allowing us to directly map the constants and variables from one to the other.

Q: Why is the energy of a solid sphere greater than that of a spherical conductor?

In a solid sphere, charge is forced to be closer together throughout the volume, which requires more work against their mutual repulsion compared to a conductor where charges spread out maximally on the surface.

Question

(a) If the same charge of $Q$ as in part (a) above is given to a spherical conductor of the same radius $R$, what will be the energy of the system?
(b) A charge of $Q$ is uniformly distributed over a spherical volume of radius $R$. Obtain an expression for the energy of the system.
Assume the earth to be a sphere of uniform mass density. Calculate this energy, given the product of the mass and the radius of the earth to be $2.5 	imes 10^{31} 	ext{ kg-m}$.
(c) What will be the corresponding expression for the energy needed to completely disassemble the planet earth against the gravitational pull amongst its constituent particles?

Sigma Percentile · Accepted Answer

### The Energy of a Spherical Conductor

Imagine you have a spherical conductor and you hand it a charge $Q$. Because it's a conductor, the charges are free to move. They despise each other, so they push each other as far away as possible, ending up entirely on the outer surface.

This means the electric field inside the conductor is absolutely zero. All the action, and therefore all the energy, is stored in the electric field outside the sphere.

To find this energy, we integrate the energy density, $u = \frac{1}{2}\epsilon_0 E^2$, from the surface of the sphere to infinity. Since the electric field outside is $E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$, the integration yields the classic result:
$$U = \frac{Q^2}{8\pi\epsilon_0 R}$$

### The Solid Sphere: Energy Inside and Out

Now, let's change the scenario. What if the charge $Q$ is uniformly distributed throughout the entire volume of a solid, non-conducting sphere?

In this case, the electric field exists both inside and outside the sphere. Inside, the field grows linearly with distance from the center: $E_{in} = \frac{Qr}{4\pi\epsilon_0 R^3}$. Outside, it behaves exactly like a point charge: $E_{out} = \frac{Q}{4\pi\epsilon_0 r^2}$.

To find the total energy, we must calculate the energy stored in both regions. Let's start with the inside. We consider a thin spherical shell of radius $r$ and thickness $dr$. The volume of this shell is $dV = 4\pi r^2 dr$.

The energy stored inside is the integral of the energy density over the volume from the center to the surface:
$$U_{in} = \int_0^R \frac{1}{2}\epsilon_0 E_{in}^2 dV = \int_0^R \frac{1}{2}\epsilon_0 \left(\frac{Qr}{4\pi\epsilon_0 R^3}ight)^2 4\pi r^2 dr = \frac{Q^2}{40\pi\epsilon_0 R}$$

Next, we calculate the energy stored outside. Since the electric field outside is identical to that of the spherical conductor, the energy stored from the surface to infinity is exactly the same:
$$U_{out} = \frac{Q^2}{8\pi\epsilon_0 R}$$

Adding these two components gives us the total self-energy of the uniformly charged solid sphere:
$$U_{total} = U_{in} + U_{out} = \frac{Q^2}{40\pi\epsilon_0 R} + \frac{Q^2}{8\pi\epsilon_0 R} = \frac{3Q^2}{20\pi\epsilon_0 R}$$

### The Gravitational Analogy: Disassembling the Earth

Now, let's shift our focus from electrostatics to gravitation. The mathematical structure of a uniform mass is identical to that of a uniform charge!

We can simply replace the electrostatic constant $\frac{1}{4\pi\epsilon_0}$ with the gravitational constant $G$, and the charge $Q$ with mass $M$. Because gravity is an attractive force, the self-energy is negative.

The gravitational self-energy of the Earth is:
$$U_g = -\frac{3GM^2}{5R}$$

To completely disassemble the Earth—to pull every constituent particle infinitely far apart against their mutual gravitational attraction—we must supply an amount of energy equal to the magnitude of its self-energy:
$$E = |U_g| = \frac{3GM^2}{5R}$$

### The Final Calculation

Finally, let's calculate this mind-boggling amount of energy. We know that the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$.

This allows us to express the gravitational constant as $G = \frac{gR^2}{M}$. Substituting this into our energy expression simplifies it beautifully:
$$E = \frac{3}{5} \left(\frac{gR^2}{M}ight) \frac{M^2}{R} = \frac{3}{5} MgR$$

We are given the product of the Earth's mass and radius: $MR = 2.5 	imes 10^{31} 	ext{ kg}\cdot	ext{m}$. Taking $g = 10 	ext{ m/s}^2$, we can plug in the values:
$$E = \frac{3}{5} 	imes 10 	imes 2.5 	imes 10^{31} = 1.5 	imes 10^{32} 	ext{ J}$$

This is the immense energy holding our planet together!

Visualized Solution

$U = \frac{Q ^{2}}{8 π ϵ _{0} R}$

$E_{in} = \frac{Q r}{4 π ϵ _{0} R ^{3}}$

$U_{in} = \frac{Q ^{2}}{40 π ϵ _{0} R}$

$U_{o u t} = \frac{Q ^{2}}{8 π ϵ _{0} R}$

$U_{t o t a l} = \frac{3 Q ^{2}}{20 π ϵ _{0} R}$

$E = \frac{3 G M ^{2}}{5 R}$

$E = 1.5 \times 1 0^{32} J$

The Sigma Insight: Electrostatic Potential Energy

The Energy of a Spherical Conductor

The Solid Sphere

The Gravitational Analogy

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U=8πϵ0​RQ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ein​=4πϵ0​R3Qr​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Uin​=40πϵ0​RQ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Uout​=8πϵ0​RQ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Utotal​=20πϵ0​R3Q2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=5R3GM2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=1.5×1032 J