IPhO 2010 · Theory problem 1

Image of a charge in a metallic object

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Introduction – Method of images

A point charge \(q\) is placed in the vicinity of a grounded metallic sphere of radius \(R\) [see Fig. 1(a)], and consequently a surface charge distribution is induced on the sphere. To calculate the electric field and potential from the distribution of the surface charge is a formidable task. However, the calculation can be considerably simplified by using the so called method of images. In this method, the electric field and potential produced by the charge distributed on the sphere can be represented as an electric field and potential of a single point charge \(q'\) placed inside the sphere (you do not have to prove it). Note: The electric field of this image charge \(q'\) reproduces the electric field and the potential only outside the sphere (including its surface).

Figure 1

Fig 1. (a) A point charge \(q\) in the vicinity of a grounded metallic sphere. (b) The electric field of the charge induced on the sphere can be represented as electric field of an image charge \(q'\).

Task 1 – The image charge

The symmetry of the problem dictates that the charge \(q'\) should be placed on the line connecting the point charge \(q\) and the center of the sphere [see Fig. 1(b)].

a) What is the value of the potential on the sphere? (0.3 points)

b) Express \(q'\) and the distance \(d'\) of the charge \(q'\) from the center of the sphere, in terms of \(q\), \(d\), and \(R\). (1.9 points)

c) Find the magnitude of force acting on charge \(q\). Is the force repulsive? (0.5 points)

Task 2 – Shielding of an electrostatic field

Consider a point charge \(q\) placed at a distance \(d\) from the center of a grounded metallic sphere of radius \(R\). We are interested in how the grounded metallic sphere affects the electric field at point \(A\) on the opposite side of the sphere (see Fig. 2). Point \(A\) is on the line connecting charge \(q\) and the center of the sphere; its distance from the point charge \(q\) is \(r\).

a) Find the vector of the electric field at point \(A\). (0.6 points)

b) For a very large distance \(r\gg d\), find the expression for the electric field by using the approximation \((1+a)^{-2}\approx 1-2a\), where \(a\ll 1\). (0.6 points)

c) In which limit of \(d\) does the grounded metallic sphere screen the field of the charge \(q\) completely, such that the electric field at point \(A\) is exactly zero? (0.3 points)

Figure 2

Fig 2. The electric field at point \(A\) is partially screened by the grounded sphere.

Task 3 – Small oscillations in the electric field of the grounded metallic sphere

A point charge \(q\) with mass \(m\) is suspended on a thread of length \(L\) which is attached to a wall, in the vicinity of the grounded metallic sphere. In your considerations, ignore all electrostatic effects of the wall. The point charge makes a mathematical pendulum (see Fig. 3). The point at which the thread is attached to the wall is at a distance \(l\) from the center of the sphere. Assume that the effects of gravity are negligible.

a) Find the magnitude of the electric force acting on the point charge \(q\) for a given angle \(\alpha\) and indicate the direction in a clear diagram (0.8 points)

b) Determine the component of this force acting in the direction perpendicular to the thread in terms of \(l\), \(L\), \(R\), \(q\) and \(\alpha\). (0.8 points)

c) Find the frequency for small oscillations of the pendulum. (1.0 points)

Figure 3

Fig 3. A point charge in the vicinity of a grounded sphere oscillates as a pendulum.

Task 4 – The electrostatic energy of the system

For a distribution of electric charges it is important to know the electrostatic energy of the system. In our problem (see Fig. 1a), there is an electrostatic interaction between the external charge \(q\) and the induced charges on the sphere, and there is an electrostatic interaction among the induced charges on the sphere themselves. In terms of the charge \(q\), radius of the sphere \(R\) and the distance \(d\) determine the following electrostatic energies:

a) the electrostatic energy of the interaction between charge \(q\) and the induced charges on the sphere; (1.0 points)

b) the electrostatic energy of the interaction among the induced charges on the sphere; (1.2 points)

c) the total electrostatic energy of the interaction in the system. (1.0 points)

Hint: There are several ways of solving this problem:

(1) In one of them, you can use the following integral,

\[ \int_d^\infty \frac{x\,dx}{(x^2-R^2)^2} =\frac{1}{2}\frac{1}{d^2-R^2}. \]

(2) In another one, you can use the fact that for a collection of \(N\) charges \(q_i\) located at points \(\vec r_i,i=1,\ldots,N\), the electrostatic energy is a sum over all pairs of charges:

\[ V=\frac{1}{2}\sum_{i=1}^{N} \sum_{\substack{j=1\\i\ne j}}^{N} \frac{1}{4\pi\varepsilon_0} \frac{q_iq_j}{|\vec r_i-\vec r_j|}. \]

Official English text of the problem.

Answers

Write \(k=1/(4\pi\varepsilon_0)\). In Task 2, let \(\hat{\mathbf e}\) point from \(q\) toward \(A\). For Task 3, define \[ d_\alpha=\sqrt{l^2+L^2-2lL\cos\alpha}, \qquad D_0=(l-L)^2-R^2, \] with \(l-L>R\) for the depicted exterior equilibrium. Take the signed perpendicular force as positive toward increasing \(\alpha\).

  • 1(a) \[ \boxed{\phi_{\mathrm{sphere}}=0.} \]

  • 1(b) \[ \boxed{q'=-\frac{R}{d}q,\qquad d'=\frac{R^2}{d}.} \] The image lies between the center and the surface nearest \(q\).

  • 1(c) \[ \boxed{F=\frac{kq^2Rd}{(d^2-R^2)^2}.} \] The force is attractive, toward the sphere’s center.

  • 2(a) \[ \boxed{ \mathbf E_A=kq\left[ \frac{1}{r^2} -\frac{R}{d\left(r-d+R^2/d\right)^2} \right]\hat{\mathbf e}. } \]

  • 2(b) For \(r\gg d\), through the first correction: \[ \boxed{ \mathbf E_A\simeq \frac{kq}{r^2} \left[ 1-\frac{R}{d} -\frac{2R}{r}\left(1-\frac{R^2}{d^2}\right) \right]\hat{\mathbf e}. } \] The leading term is \(kq(1-R/d)\hat{\mathbf e}/r^2\).

  • 2(c) \[ \boxed{d\to R^+\quad\Longrightarrow\quad \mathbf E_A\to\mathbf 0.} \]

  • 3(a) \[ \boxed{ F(\alpha)= \frac{kq^2R\sqrt{l^2+L^2-2lL\cos\alpha}} {\left(l^2+L^2-2lL\cos\alpha-R^2\right)^2}. } \] Draw the force arrow from the charge directly toward the sphere’s center.

  • 3(b) \[ \boxed{ F_\perp= -\frac{kq^2Rl\sin\alpha} {\left(l^2+L^2-2lL\cos\alpha-R^2\right)^2}. } \]

  • 3(c) The ordinary frequency and angular frequency are \[ \boxed{ f=\frac{1}{2\pi}\sqrt{\frac{kq^2Rl}{mL D_0^2}}, \qquad \omega_0=2\pi f. } \]

  • 4(a) Charge–surface interaction: \[ \boxed{U_{q\sigma}=-\frac{kq^2R}{d^2-R^2}.} \]

  • 4(b) Mutual energy of the induced charges: \[ \boxed{U_{\sigma\sigma}=+\frac{kq^2R}{2(d^2-R^2)}.} \]

  • 4(c) Total interaction energy: \[ \boxed{U_{\mathrm{tot}}=-\frac{kq^2R}{2(d^2-R^2)}.} \]

Solution

1(a)

A conductor in electrostatic equilibrium is an equipotential. Grounding fixes its potential to that of Earth, taken as zero: \[ \boxed{\phi_{\mathrm{sphere}}=0.} \]

1(b)

Let \(O\) be the center. Measure axial positions from \(O\) toward the external charge, so that \(q\) is at \(d>R\) and the image is at \(d'\), inside the sphere.

At the surface point nearest \(q\), the zero-potential condition gives \[ k\left(\frac{q}{d-R}+\frac{q'}{R-d'}\right)=0. \] At the opposite surface point, \[ k\left(\frac{q}{d+R}+\frac{q'}{R+d'}\right)=0. \] Consequently, \[ \frac{R-d'}{d-R}=\frac{R+d'}{d+R}. \] Cross-multiplication yields \[ (R-d')(d+R)=(R+d')(d-R), \] hence \[ \boxed{d'=\frac{R^2}{d}}. \] Substituting back, \[ \boxed{q'=-\frac{R}{d}q}. \]

These values also satisfy the boundary condition everywhere on the surface. For a surface point whose radius makes angle \(\theta\) with the direction toward \(q\), let its distances from \(q\) and \(q'\) be \(s\) and \(s'\). Then \[ s^2=d^2+R^2-2dR\cos\theta, \] and \[ s'^2=d'^2+R^2-2d'R\cos\theta =\frac{R^2}{d^2}s^2. \] Thus \(s'=(R/d)s\), and \[ k\left(\frac{q}{s}+\frac{q'}{s'}\right)=0. \]

1(c)

At the external charge, the image reproduces the field of the induced charges. The separation between \(q\) and its image is \[ d-d'=\frac{d^2-R^2}{d}. \] Coulomb’s law therefore gives \[ F=\frac{k|qq'|}{(d-d')^2} =\frac{kq^2(R/d)}{\left[(d^2-R^2)/d\right]^2}. \] Hence \[ \boxed{F=\frac{kq^2Rd}{(d^2-R^2)^2}}. \]

Since \(q'\) has the opposite sign to \(q\), the force is always attractive. It points along the line from \(q\) toward \(O\).

2(a)

Choose the positive axis from \(q\) toward \(A\), with unit vector \(\hat{\mathbf e}\). Put the sphere’s center at \(x=0\). Then \[ x_q=-d,\qquad x_{q'}=-d',\qquad x_A=r-d. \] Because \(A\) is outside the sphere on the opposite side, \(r>d+R\).

The distances from \(A\) to the two charges are \[ r,\qquad r-d+d'=r-d+\frac{R^2}{d}. \] By superposition, \[ \mathbf E_A= k\left[ \frac{q}{r^2} +\frac{q'}{(r-d+d')^2} \right]\hat{\mathbf e}. \] Therefore \[ \boxed{ \mathbf E_A=kq\left[ \frac{1}{r^2} -\frac{R}{d\left(r-d+R^2/d\right)^2} \right]\hat{\mathbf e}. } \] The sign of the scalar coefficient determines the direction along the axis.

2(b)

Define \[ \delta=d-\frac{R^2}{d}. \] Since \(0<\delta<d\), the condition \(r\gg d\) implies \(\delta/r\ll1\). We can write \[ \mathbf E_A= \frac{kq}{r^2} \left[ 1-\frac{R}{d}\left(1-\frac{\delta}{r}\right)^{-2} \right]\hat{\mathbf e}. \]

Using the supplied approximation with \(a=-\delta/r\), \[ \left(1-\frac{\delta}{r}\right)^{-2} \simeq1+\frac{2\delta}{r}. \] It follows that \[ \mathbf E_A\simeq \frac{kq}{r^2} \left[ 1-\frac{R}{d} -\frac{2R\delta}{dr} \right]\hat{\mathbf e}. \] Substituting \(\delta\), \[ \boxed{ \mathbf E_A\simeq \frac{kq}{r^2} \left[ 1-\frac{R}{d} -\frac{2R}{r}\left(1-\frac{R^2}{d^2}\right) \right]\hat{\mathbf e}. } \]

To leading order, the distant field is that of the effective net charge \[ q+q'=q\left(1-\frac{R}{d}\right). \]

2(c)

As the external charge approaches the sphere, \[ d\to R^+,\qquad d'\to R,\qquad q'\to-q. \] Thus the charge and its image approach the same position with equal and opposite charges. Their fields cancel at every fixed exterior observation point away from the contact point.

In particular, the exact expression in 2(a) gives \[ \boxed{\lim_{d\to R^+}\mathbf E_A=\mathbf 0}. \] This is a limit with the charge approaching the surface from outside.

3(a)

Treat the conductor as remaining in electrostatic equilibrium as the charge moves. Let \(P\) be the thread’s attachment point and \(Q\) the charge’s position. Choose \[ O=(0,0),\qquad P=(l,0). \] For the angle shown in Fig. 3, \[ Q=(l-L\cos\alpha,\;L\sin\alpha). \] Therefore the instantaneous center-to-charge distance is \[ d_\alpha=|OQ| =\sqrt{(l-L\cos\alpha)^2+L^2\sin^2\alpha} =\sqrt{l^2+L^2-2lL\cos\alpha}. \]

Applying the result of 1(c), \[ \boxed{ F(\alpha)= \frac{kq^2R\,d_\alpha}{(d_\alpha^2-R^2)^2} = \frac{kq^2R\sqrt{l^2+L^2-2lL\cos\alpha}} {\left(l^2+L^2-2lL\cos\alpha-R^2\right)^2}. } \]

The force points from \(Q\) toward \(O\).

Diagram description: Draw the sphere centered at \(O\), with \(P\) to its right on a horizontal line and \(OP=l\). Place \(Q\) above that line and to the left of \(P\), with \(PQ=L\) and \(\angle OPQ=\alpha\). Draw an arrow starting at \(Q\), along \(QO\), pointing toward \(O\), and label it \(\mathbf F_e\). Label \(OQ=d_\alpha\).

3(b)

A unit vector perpendicular to the thread, in the direction of increasing \(\alpha\), is \[ \hat{\mathbf e}_\alpha=(\sin\alpha,\cos\alpha). \] Since the force points toward \(O\), \[ \mathbf F_e=-\frac{F}{d_\alpha} (l-L\cos\alpha,\;L\sin\alpha). \] Its tangential component is \[ \begin{aligned} F_\perp &=\mathbf F_e\cdot\hat{\mathbf e}_\alpha\\ &=-\frac{F}{d_\alpha} \left[(l-L\cos\alpha)\sin\alpha +L\sin\alpha\cos\alpha\right]\\ &=-\frac{Fl\sin\alpha}{d_\alpha}. \end{aligned} \] Substitution gives \[ \boxed{ F_\perp= -\frac{kq^2Rl\sin\alpha} {\left(l^2+L^2-2lL\cos\alpha-R^2\right)^2}. } \]

For positive \(\alpha\), this component acts toward decreasing \(\alpha\), so it is restoring. Its magnitude is obtained by replacing \(-\sin\alpha\) with \(|\sin\alpha|\).

3(c)

At the equilibrium \(\alpha=0\), \[ d_0=l-L. \] The depicted equilibrium requires \(l-L>R\), so the charge lies outside the sphere.

Tension has no component perpendicular to the thread. Newton’s second law in the tangential direction is therefore \[ mL\ddot\alpha=F_\perp. \] For small \(\alpha\), \[ \sin\alpha\simeq\alpha, \] and \[ d_\alpha^2-R^2 =(l-L)^2-R^2+lL\alpha^2+O(\alpha^4). \] Writing \(D_0=(l-L)^2-R^2\), the linearized equation becomes \[ mL\ddot\alpha =-\frac{kq^2Rl}{D_0^2}\alpha, \] or \[ \ddot\alpha+\omega_0^2\alpha=0, \qquad \omega_0^2=\frac{kq^2Rl}{mL D_0^2}. \]

Thus the ordinary frequency is \[ \boxed{ f=\frac{\omega_0}{2\pi} =\frac{1}{2\pi} \sqrt{ \frac{kq^2Rl} {mL\left[(l-L)^2-R^2\right]^2} }. } \]

The approximation requires sufficiently small amplitude that both \[ |\alpha|\ll1, \qquad lL\alpha^2\ll D_0 \] hold.

4(a)

Omit the external point charge’s self-energy and take the interaction energy to vanish as \(d\to\infty\).

Let \(\phi_\sigma\) denote the potential produced by the induced surface charges. At the position of \(q\), the image gives \[ \phi_\sigma(q)=\frac{kq'}{d-d'} =-\frac{kqR}{d^2-R^2}. \] The interaction energy between \(q\) and the induced charges is \[ U_{q\sigma}=q\phi_\sigma(q). \] There is no factor \(1/2\) here: each pair consisting of \(q\) and an induced charge is counted once. Hence \[ \boxed{ U_{q\sigma}=-\frac{kq^2R}{d^2-R^2}. } \]

4(b)

Let \(\sigma\) be the induced surface charge density, and let \(\phi_q\) be the potential due to the external charge. The mutual energy of the induced charges is \[ U_{\sigma\sigma} =\frac12\int_S \sigma\,\phi_\sigma\,dS. \] The factor \(1/2\) prevents double counting of pairs within the surface distribution.

On the grounded sphere, \[ \phi_q+\phi_\sigma=0. \] Therefore \[ U_{\sigma\sigma} =-\frac12\int_S\sigma\,\phi_q\,dS. \] The remaining integral equals the charge–surface interaction energy: \[ \int_S\sigma\,\phi_q\,dS =kq\int_S\frac{\sigma\,dS}{|\mathbf r-\mathbf r_q|} =q\phi_\sigma(q) =U_{q\sigma}. \] Thus \[ U_{\sigma\sigma}=-\frac12U_{q\sigma}, \] giving \[ \boxed{ U_{\sigma\sigma}=+\frac{kq^2R}{2(d^2-R^2)}. } \]

4(c)

Adding the two physical contributions, \[ U_{\mathrm{tot}}=U_{q\sigma}+U_{\sigma\sigma} =\frac12U_{q\sigma}. \] Consequently, \[ \boxed{ U_{\mathrm{tot}}=-\frac{kq^2R}{2(d^2-R^2)}. } \]

This also follows from the work required to bring \(q\) quasistatically from infinity: \[ U_{\mathrm{tot}}(d) =-\int_d^\infty \frac{kq^2R\,x}{(x^2-R^2)^2}\,dx =-\frac{kq^2R}{2(d^2-R^2)}. \]

As a consistency check, the force in the direction of increasing \(d\) is \[ -\frac{dU_{\mathrm{tot}}}{dd} =-\frac{kq^2Rd}{(d^2-R^2)^2}, \] which reproduces the attractive force found in 1(c).

The solution was checked against the official solution.