Question

The minimum strength of a uniform electric field which can tear a conducting uncharged thin-walled sphere into two parts is known to be $E_0$.

Determine the minimum electric field strength $E_1$ required to tear the sphere of twice as large radius if the thickness of its walls is the same as in the former case.

Answer 1

$\frac{E_0}{\sqrt{2}}$

Answer 2

1. Electric Field and Surface Charge Density: For a conducting sphere of radius $R$ placed in a uniform external electric field $E$, the total electric field at the surface is radial and its magnitude is $E_{total}(R) = 3E \cos\theta$, where $\theta$ is the polar angle measured from the direction of the external field. The surface charge density is given by $\sigma = \epsilon_0 E_{total}(R) = 3\epsilon_0 E \cos\theta$.

2. Electric Pressure: The force per unit area (pressure) acting outwards on the surface of the conductor is given by the product of the surface charge density and the total electric field at the surface: $P = \sigma E_{total}(R) = (3\epsilon_0 E \cos\theta)(3E \cos\theta) = 9\epsilon_0 E^2 \cos^2\theta$.

3. Force on a Hemisphere: To find the force that tears the sphere into two parts, we consider the force acting on one hemisphere. This force is obtained by integrating the pressure over the surface area of the hemisphere. For the top hemisphere ($0 \le \theta \le \pi/2$), the force $F_{hemi}$ is given by: $F_{hemi} = \int_{0}^{2\pi} \int_{0}^{\pi/2} P \, dA = \int_{0}^{2\pi} \int_{0}^{\pi/2} (9\epsilon_0 E^2 \cos^2\theta) (R^2 \sin\theta \, d\theta \, d\phi)$ Evaluating this integral gives $F_{hemi} = 6\pi\epsilon_0 E^2 R^2$.

4. Tensile Stress: This force $F_{hemi}$ is resisted by the material of the thin-walled sphere at the equator. The cross-sectional area of the material at the equator is approximately $A = 2\pi R t$, where $t$ is the wall thickness. The tensile stress $\tau$ in the material is the force per unit area: $\tau = \frac{F_{hemi}}{A} = \frac{6\pi\epsilon_0 E^2 R^2}{2\pi R t} = \frac{3\epsilon_0 E^2 R}{t}$

5. Tearing Condition: The sphere tears when this tensile stress $\tau$ exceeds the material's tensile strength, $S$. So, the minimum electric field $E_0$ required to tear the first sphere is related to $S$ by: $S = \frac{3\epsilon_0 E_0^2 R}{t}$

6. Second Sphere: For the second sphere with twice the radius ($R' = 2R$) and the same wall thickness $t$, the minimum electric field required is $E_1$. The stress in this sphere is: $\tau' = \frac{3\epsilon_0 E_1^2 R'}{t} = \frac{3\epsilon_0 E_1^2 (2R)}{t}$ The tearing condition is $\tau' = S$: $\frac{6\epsilon_0 E_1^2 R}{t} = \frac{3\epsilon_0 E_0^2 R}{t}$

7. Result: Simplifying the equation, we get $6E_1^2 = 3E_0^2$, which leads to $E_1^2 = \frac{1}{2}E_0^2$. Therefore, $E_1 = \frac{E_0}{\sqrt{2}}$.