P3.7

• There are two charges +q at (0,0, 3d) -2q at (0, 0, d). The xy plane is a grounded conductor. Find the force on the +q charge.

  Answer:

  Method of images Grounded plane Superposition

  Setup. Real charges: \(+q\) at \((0,0,3d)\) and \(-2q\) at \((0,0,d)\). The grounded conducting plane is the \(xy\)-plane (\(z=0\)). For a grounded plane, each real charge has an image of equal magnitude and opposite sign mirrored through the plane:

  • \(+q\) at \(z=3d \;\Rightarrow\) image \(-q\) at \(z=-3d\).
  • \(-2q\) at \(z=d \;\Rightarrow\) image \(+2q\) at \(z=-d\).

  The electric field in the region \(z>0\) equals the field of the two real charges plus these image charges (with the conductor removed). The force on the \(+q\) charge is then just Coulomb’s law with those other three sources.

  Forces on \(+q\) at \(z=3d\) (all along \(\hat{\mathbf z}\))

  \[ k=\frac{1}{4\pi\varepsilon_0},\qquad \Delta z = |z_1-z_2|. \]

  • From real \(-2q\) at \(z=d\): \(\Delta z=2d\), attractive (toward \(-z\)): \[ \mathbf F_1 = -\,k\,\frac{2q^2}{(2d)^2}\,\hat{\mathbf z} = -\,k\,\frac{q^2}{2d^2}\,\hat{\mathbf z}. \]
  • From image \(+2q\) at \(z=-d\): \(\Delta z=4d\), repulsive (toward \(+z\)): \[ \mathbf F_2 = +\,k\,\frac{2q^2}{(4d)^2}\,\hat{\mathbf z} = +\,k\,\frac{q^2}{8d^2}\,\hat{\mathbf z}. \]
  • From image \(-q\) at \(z=-3d\): \(\Delta z=6d\), attractive (toward \(-z\)): \[ \mathbf F_3 = -\,k\,\frac{q^2}{(6d)^2}\,\hat{\mathbf z} = -\,k\,\frac{q^2}{36d^2}\,\hat{\mathbf z}. \]

  Result

  \[ \mathbf F = \mathbf F_1+\mathbf F_2+\mathbf F_3 = k\,\frac{q^2}{d^2}\left(\;-\frac{1}{2}+\frac{1}{8}-\frac{1}{36}\;\right)\hat{\mathbf z} = -\,\frac{29}{72}\,k\,\frac{q^2}{d^2}\,\hat{\mathbf z}. \]

  Direction: toward the plane (negative \(z\)). The downward attractions from the nearby \(-2q\) and the image \(-q\) dominate over the upward repulsion from the image \(+2q\).

P3.10

• In ex.3.2 (below) we assumed that the conducting sphere was grounded (V = 0). But with the addition of a second image charge, the same basic model will handle the case of a second image charge, the same basic model will handle the case of a sphere at any potential V_0 (relative, of course, to infinity). What charge should you use, and where should you put it? Find the force of attraction between a point charge q and a neutral conducting sphere.
  
  3.2]In one sentence, justify Earnshaw's theorem: a charged particle cannot be held in stable equilibrium (in otherwise empty space) by electrostatic forces alone. As an example, consider a cubicle arrangement of eight fixed charges. It looks, off hand, as though a positive charge at the center would be suspended in midair since it is repelled away from each corner. Where is the leak in this "electrostatic bottle?

  Answer:

  Method of images Sphere at arbitrary potential Neutral sphere force

  Geometry. A conducting sphere of radius \(a\) is centered at the origin. A point charge \(q\) is on the \(+z\)-axis at \(r=d\) with \(d>a\). The standard grounded-sphere image is \[ q'=-\frac{a}{d}\,q\quad\text{located at}\quad r'=\frac{a^{2}}{d}\ \ (\text{on the }+z\text{ axis}). \] This pair (\(q,q'\)) makes the potential on \(r=a\) vanish.

  Sphere at any specified potential \(V_0\)

  Add a second image at the center, \(Q_0\) at \(r=0\). A point charge at the center contributes a constant potential on the surface: \(V_{\text{surf}}=k Q_0/a\) (with \(k=\tfrac{1}{4\pi\varepsilon_0}\)). Since the \((q,q')\) pair already yields \(V=0\) on the surface, choose \[ \boxed{\,Q_0 = 4\pi\varepsilon_0\,a\,V_0\,}\quad\text{(placed at the center).} \] This sets the conductor’s potential to \(V_0\) while preserving the correct exterior field.

  Neutral conducting sphere

  Neutrality requires the total induced charge on the sphere to be zero. In the image picture, that total equals the sum of image charges. Hence set \[ \boxed{\,Q_0 = -\,q' = \frac{a}{d}\,q\,}\quad\Rightarrow\quad V_{\text{surf}} = k\,\frac{Q_0}{a} = k\,\frac{q}{d}. \] (The sphere "floats" to potential \(kq/d\).)

  Force on \(q\) due to the (neutral) sphere

  The force on the real charge is just the vector sum of forces from the two images inside the sphere (the conductor itself exerts no additional force in the image method):

  • From \(q'\) at \(r'\): separation \(d-r' = d - \tfrac{a^{2}}{d}\); force along \(-\hat{\mathbf z}\) (attractive), magnitude \(k\,|q q'|/(d-r')^{2}\).
  • From \(Q_0\) at the center: separation \(d\); force along \(+\hat{\mathbf z}\) (repulsive if \(q>0\)), magnitude \(k\,|q Q_0|/d^{2}\).

  \[ \mathbf F = k\,\frac{q\,q'}{(d-\tfrac{a^{2}}{d})^{2}}\,\hat{\mathbf z} + k\,\frac{q\,Q_0}{d^{2}}\,\hat{\mathbf z} \quad\text{with}\quad q'=-\frac{a}{d}q,\;\; Q_0=\frac{a}{d}q. \]

  After substitution and simplification, the net force (on \(q\)) is purely radial toward the sphere’s center:

  \[ \boxed{\; \mathbf F = -\,\frac{1}{4\pi\varepsilon_0}\, \frac{q^{2}\,a^{3}\,\big(2d^{2}-a^{2}\big)} {d^{3}\,\big(d^{2}-a^{2}\big)^{2}}\;\hat{\mathbf z} \;} \qquad(d>a). \]

  Checks. For large separation \(d\gg a\): \(\displaystyle \mathbf F \sim -\,\frac{2k q^{2} a^{3}}{d^{5}}\,\hat{\mathbf z}\) (always attractive). For a grounded sphere (set \(Q_0=0\)) you recover \(\displaystyle \mathbf F_{\text{ground}}=-\,k\,\frac{q^{2} a\,d}{(d^{2}-a^{2})^{2}}\,\hat{\mathbf z}\).

  ---

  One-liner (Earnshaw, as referenced): In empty space \(\nabla^2 V=0\); any local potential extremum would require \(\nabla^2 V\) to be of the same sign, which contradicts Laplace’s equation—hence no stable electrostatic equilibrium.

P3.12

• A uniform line charge λ is placed on an infinite straight wire, a distance d above a grounded conducting plane. (ley's say the wire runs parallel to the x-axis and directly above it, and the conducting plane is the xy plane.)
  
  a)Find the potential in the region above the plane.
  b)Find the charge density

  Answer:

  Method of images Grounded plane Surface charge density

  Setup. A uniform line charge \(+\lambda\) lies along the \(x\)-axis at transverse position \((y,z)=(0,d)\) above a grounded conducting plane \(z=0\). By the method of images, the region \(z>0\) is equivalent to free space containing the real line \(+\lambda\) at \(z=d\) and an image line \(-\lambda\) at \(z=-d\). All quantities are independent of \(x\).

  (a) Potential in the region \(z>0\)

  Let \(r_{+}=\sqrt{y^{2}+(z-d)^{2}}\) be the distance to \(+\lambda\) and \(r_{-}=\sqrt{y^{2}+(z+d)^{2}}\) the distance to \(-\lambda\). A line charge contributes \( ( \lambda / 2\pi\varepsilon_{0} ) \ln r \) (up to a constant), and the grounded boundary fixes that constant so that \(V=0\) on \(z=0\). Hence

  \[ V(y,z)\;=\;\frac{\lambda}{2\pi\varepsilon_0}\,\ln\!\frac{r_{-}}{r_{+}} \;=\;\frac{\lambda}{4\pi\varepsilon_0}\, \ln\!\frac{y^{2}+(z+d)^{2}}{y^{2}+(z-d)^{2}}\,,\qquad (z>0). \]

  Check: At \(z=0\) the ratio is 1 so \(V=0\); far away \(V\to 0\) like a dipole.

  (b) Induced surface charge density on the plane

  The surface charge density on a conductor is \(\sigma = \varepsilon_0 E_n\) (with \(E_n\) the normal component just outside the surface). Here \(E_z=-\partial V/\partial z\), so at \(z=0^+\):

  \[ \frac{\partial V}{\partial z} = \frac{\lambda}{4\pi\varepsilon_0}\left[ \frac{2(z+d)}{y^{2}+(z+d)^{2}} - \frac{2(z-d)}{y^{2}+(z-d)^{2}} \right] \xrightarrow[z\to 0^+]{}\; \frac{\lambda}{4\pi\varepsilon_0}\,\frac{4d}{y^{2}+d^{2}}. \]

  \[ \boxed{\;\sigma(y)=\varepsilon_0 E_z(0^+) = -\,\varepsilon_0\,\frac{\partial V}{\partial z}\Big|_{0^+} = -\,\frac{\lambda\,d}{\pi\,(y^{2}+d^{2})}\;} \quad\text{(independent of \(x\))}. \]

  Sanity checks: \(\sigma(y)\le 0\) (negative under a positive line), it peaks at \(y=0\): \(\sigma(0)=-\lambda/(\pi d)\), and \(\displaystyle \int_{-\infty}^{\infty}\sigma(y)\,dy = -\lambda\) (total induced charge per unit length equals \(-\lambda\)).

P3.13

• Two semi-infinite grounded conducting planes meet at right angles. In the region between them, a point charge 𝑞 is located at position (𝑎,𝑏).
  
  Set up the image charge configuration and calculate the potential in this region.
  
  What charges are needed, and where should they be placed?
  
  What is the force on q?
  
  How much work would it take to bring q in from infinity?
  
  Suppose instead that the planes met at some angle other than 90∘. Would the method of images still work? If not, for what special angles does it work?

  Problem 3.13 — Geometry & Image Charges

  Two semi-infinite grounded conducting planes are the coordinate axes x=0 and y=0 (both at V=0). A point charge q sits in the first quadrant at (a,b). The image charges needed to enforce V=0 on both planes are shown.

  Answer:

  Problem 3.13 — Method of Images (Right Angle)

  Image charges for planes at x=0 and y=0 with real q at (a,b): −q at (−a,b), −q at (a,−b), +q at (−a,−b).

  Potential (x>0, y>0)

  $$ V(x,y)=\frac{1}{4\pi\varepsilon_0}\left[ \frac{q}{\sqrt{(x-a)^2+(y-b)^2}} -\frac{q}{\sqrt{(x+a)^2+(y-b)^2}} -\frac{q}{\sqrt{(x-a)^2+(y+b)^2}} +\frac{q}{\sqrt{(x+a)^2+(y+b)^2}} \right] $$

  Force on q

  Let \(k=1/(4\pi\varepsilon_0)\). Then

  $$F_x = -k\,\frac{q^2}{4a^2} + k\,\frac{q^2\,a}{4\,(a^2+b^2)^{3/2}}$$

  $$F_y = -k\,\frac{q^2}{4b^2} + k\,\frac{q^2\,b}{4\,(a^2+b^2)^{3/2}}$$

  Work to bring q from infinity

  $$ W=\tfrac{1}{2}q\,V_{\text{images}}(a,b) = k\,\frac{q^2}{4}\left(-\frac{1}{a}-\frac{1}{b}+\frac{1}{\sqrt{a^2+b^2}}\right) $$

  Other wedge angles

  Finite image sets close when the wedge angle is \(\alpha=\pi/n\) (integer \(n\ge2\)); otherwise use separation of variables.

P3.17

•  A rectangular pipe, running parallel to the z-axis (from −∞ to +∞), has three grounded metal sides, at y = 0, y = a, and x = 0. The fourth side, at x = b, is maintained at a specified potential V₀(y).
  
  (a) Develop a general formula for the potential inside the pipe.
  (b) Find the potential explicitly, for the case V₀(y) = V₀ (a constant).  

  Problem 3.17 — Rectangular Pipe (z-directed, infinite)

  Cross-section in the xy-plane. Three walls are grounded: x=0, y=0, and y=a. The fourth wall at x=b is held at a specified potential V₀(y). The pipe runs along z (into the page).

  Answer:

P3.18

• A cubical box (side length a) consists of five metal plates, which are welded together and grounded. The top is made of a separate sheet of metal, insulated from the others, and held at a constant potential V₀. Find the potential inside the box. [What should the potential at the center (a/2, a/2, a/2) be? Check numerically that your formula is consistent with this value.]

  Problem 3.18 — Geometry

  Cube of side a occupying 0 ≤ x,y,z ≤ a. Planes x=0, y=0, z=0 are grounded (V=0); the top face z=a is held at V = V₀.

  Answer:

P3.21

• The potential at the surface of a sphere (radius R) is given by  
  
  V₀ = k cos(3θ),
  
  where k is a constant. Find the potential inside and outside the sphere, as well as the surface charge density σ(θ) on the sphere. (Assume there’s no charge inside or outside the sphere.)

  Problem 3.21 — Spherical Boundary V(R,θ) = k cos(3θ)

  A sphere of radius R forms the boundary between an interior region (r<R) and an exterior region (r>R). The surface potential is prescribed as V(R,θ)=k cos(3θ). There are no charges in either region.

  Answer:

P3.26

• Solve Laplace’s equation by separation of variables in cylindrical coordinates, assuming there is no dependence on z (cylindrical symmetry). Make sure you find all solutions to the radial equation; in particular, your result must accommodate the case of an infinite line charge, for which (of course) we already know the answer.

  Problem 3.26 — Cylindrical Coordinates Setup

  Laplace’s equation in cylindrical coordinates, independent of z. The geometry is symmetric around the z-axis, with a line charge (or general boundary conditions) defining \(V(r,\phi)\).

  Answer: