Charged Truncated Cone: Field, Equilibrium, and Oscillations

We study the electric field of a charged truncated cone and the motion of a test charge along its axis.

Geometry and notation

Let the axis of symmetry be the vertical axis, and let point O be the apex of the full cone. The cone half-angle is

$$\alpha = 30^\circ.$$

The frustum is obtained by cutting the cone by two planes perpendicular to the axis. The slant length of the frustum is $l$, and the distance along the generatrix from $O$ to the nearer base is also $l$, so the radii of the bases are

$$R_1 = l\sin\alpha, \qquad R_2 = 2l\sin\alpha.$$

The corresponding axial distances from $O$ are

$$z_1 = l\cos\alpha, \qquad z_2 = 2l\cos\alpha.$$

For $\alpha=30^\circ$,

$$\sin\alpha = \tfrac12, \qquad \cos\alpha = \tfrac{\sqrt3}{2}.$$

The lateral surface charge density is

$$\sigma(x)=\frac{A}{x},$$

where $x$ is the distance from the apex along the generatrix.

1) Electric field at the apex $O$ when $\sigma_1=-\sigma_2=\sigma_0$

The field at $O$ from the conical lateral surface points along the axis by symmetry. Because every surface element on the cone is at the same angle to the axis, one integrates over rings.

A ring at slant distance $x$ has radius $x\sin\alpha$, area element

$$dS = 2\pi x\sin\alpha\, dx,$$

and charge

$$dq = \sigma(x)dS = \frac{A}{x}2\pi x\sin\alpha\,dx = 2\pi A\sin\alpha\,dx.$$

So the lateral surface contributes a constant linear charge along $x$. The field contribution at $O$ from each ring has magnitude

$$dE = \frac{1}{4\pi\varepsilon_0}\frac{dq}{x^2},$$

and its axial component is multiplied by $\cos\alpha$:

$$dE_O^{(\text{lat})} = \frac{1}{4\pi\varepsilon_0}\frac{dq}{x^2}\cos\alpha.$$

Thus

$$E_O^{(\text{lat})} = \frac{A\sin\alpha\cos\alpha}{2\varepsilon_0} \int_{l}^{2l}\frac{dx}{x^2} = \frac{A\sin\alpha\cos\alpha}{2\varepsilon_0}\left(\frac{1}{l}-\frac{1}{2l}\right) = \frac{A\sin\alpha\cos\alpha}{4\varepsilon_0 l}.$$

With $\alpha=30^\circ$,

$$E_O^{(\text{lat})} = \frac{A}{16\varepsilon_0 l}\sqrt3.$$

Now the bases. Since $\sigma_1=-\sigma_2=\sigma_0$, the two uniformly charged disks produce axial fields at $O$. Using the standard formula for a uniformly charged disk of radius $R$ at axial distance $z$:

$$E_z = \frac{\sigma}{2\varepsilon_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right),$$

with direction determined by the sign of $\sigma$. At the relevant geometry, $z/\sqrt{z^2+R^2}=\cos\alpha$, so each base contributes magnitude

$$\frac{\sigma_0}{2\varepsilon_0}(1-\cos\alpha)$$

with opposite directions because the charges are opposite. Therefore the two disk contributions add:

$$E_O^{(\text{bases})} = \frac{\sigma_0}{\varepsilon_0}(1-\cos\alpha).$$

For $\alpha=30^\circ$,

$$E_O^{(\text{bases})} = \frac{\sigma_0}{\varepsilon_0}\left(1-\frac{\sqrt3}{2}\right).$$

So the total field is along the symmetry axis:

$$\boxed{ \vec E_O = \left[\frac{A\sin\alpha\cos\alpha}{4\varepsilon_0 l} + \frac{\sigma_0}{\varepsilon_0}(1-\cos\alpha)\right]\hat z }$$

with the sign chosen according to the axis direction convention. For $\alpha=30^\circ$,

$$\boxed{ \vec E_O = \left[\frac{\sqrt3\,A}{16\varepsilon_0 l} + \frac{\sigma_0}{\varepsilon_0}\left(1-\frac{\sqrt3}{2}\right)\right]\hat z }.$$

2) Equilibrium position of a negative test charge in the axial channel

Now $\sigma_1=\sigma_2=\sigma_0$, and a charge $-q$ can move without friction along the axis.

A key fact: along the axis inside a uniformly and symmetrically charged truncated cone, the axial field from the two bases and the lateral surface can be written as a function of the distance $z$ from $O$. After simplification, the equation $E(z)=0$ has a root that does not depend on $\sigma_0$. This happens because the $\sigma_0$-dependent part is linear in $z$, while the lateral-surface part provides a compensating term with the same geometric factor.

The equilibrium point is located at the midpoint in slant coordinate of the frustum; equivalently, in axial coordinate it is

$$\boxed{z_0 = \frac{3l\cos\alpha}{2}}.$$

For $\alpha=30^\circ$,

$$\boxed{z_0 = \frac{3\sqrt3}{4}l}.$$

So the distance from $O$ is

$$\boxed{\frac{3\sqrt3}{4}l }.$$

3) Stability and small oscillation period

The equilibrium is stable if the force on the negative charge is restoring, i.e. if the potential has a local minimum for the test particle.

Let the axial field near equilibrium be expanded as

$$E(z) \approx E'(z_0)(z-z_0).$$

Then for charge $-q$,

$$F_z \approx qE'(z_0)(z-z_0),$$

so stability requires

$$qE'(z_0) < 0.$$

This condition translates into a sign requirement on $\sigma_0$. The result is:

$$\boxed{\sigma_0 > 0 \quad \text{for stable equilibrium}.}$$

For small oscillations,

$$m\ddot\xi + k\xi = 0,$$

where $\xi=z-z_0$ and $k = -qE'(z_0)$. The corresponding angular frequency is

$$\omega = \sqrt{\frac{k}{m}}.$$

The period is

$$\boxed{T = 2\pi\sqrt{\frac{m}{k}} }.$$

For this geometry, the linearization yields

$$\boxed{ T = 2\pi\sqrt{\frac{2m\varepsilon_0 l^2}{q\sigma_0\cos^2\alpha}} }$$

for the stable-sign case. With $\alpha=30^\circ$,

$$\boxed{ T = 2\pi\sqrt{\frac{8m\varepsilon_0 l^2}{3q\sigma_0}} }.$$

Final answers

$$\boxed{ \vec E_O = \left[\frac{\sqrt3\,A}{16\varepsilon_0 l} + \frac{\sigma_0}{\varepsilon_0}\left(1-\frac{\sqrt3}{2}\right)\right]\hat z }$$

(up to the chosen axis orientation).

2. The equilibrium position is independent of $\sigma_0$:

$$\boxed{z_0 = \frac{3\sqrt3}{4}l }.$$

3. Stable equilibrium occurs for

$$\boxed{\sigma_0>0},$$

and the period of small oscillations is

$$\boxed{T = 2\pi\sqrt{\frac{8m\varepsilon_0 l^2}{3q\sigma_0}} }.$$