Physics
Exchange of dominance between monopole and induced dipole fields of charged metal sphere
Step-by-step physics solution: Exchange of dominance between monopole and induced dipole fields of charged metal sphere
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1. What the problem is asking
We have two almost identical arrangements of charges
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Diagram (b) – three point charges: two identical charges +q placed symmetrically about point R, and a third charge +2q a little farther away (the “top‑right” charge).
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Diagram (c) – the two +q charges are unchanged, but the +2q point charge is replaced by a conducting sphere that carries the same total charge +2q.
Point R is exactly halfway between the two +q charges, so the electric field produced by those two charges cancels at R.
The question: Is the magnitude of the electric field at R larger, smaller, or the same when the +2q point charge is replaced by the charged conducting sphere?
2. Step‑by‑step solution
Step 1 – Write the field at R for diagram (b)
Because the two +q charges are equal and opposite in position with respect to R, their contributions cancel:
[ \mathbf E_{+q}^{(\text{left})}(R)+\mathbf E_{+q}^{(\text{right})}(R)=\mathbf 0 . ]
Therefore the only contribution is the field of the single point charge +2q.
If the distance from that charge to R is (r),
[ \boxed{E_{(b)} = \frac{k\,2q}{r^{2}}}\qquad\text{(directed away from the +2q charge)} . ]
(k=1/4\pi\varepsilon_{0}) is Coulomb’s constant.
Step 2 – What changes when the point charge becomes a conducting sphere?
A conducting sphere with total charge +2q still produces the same monopole (overall) field as a point charge as long as we look at points outside the sphere.
Mathematically, for any point whose distance from the centre of the sphere is (R_{\text{out}} > a) (where (a) is the sphere radius),
[ \mathbf E_{\text{monopole}}(R_{\text{out}})=\frac{k\,2q}{R_{\text{out}}^{2}}\;\hat{\mathbf r}, ]
exactly the same expression we used in diagram (b).
However, the external +q charges polarise the sphere. Positive charge is repelled to the side opposite the +q’s, leaving a relative deficit of positive charge on the near side. The resulting charge distribution can be represented (to first order) by a dipole moment (\mathbf p) that points away from the two +q charges.
Consequences:
- The monopole term still points away from the sphere’s centre (same direction as in (b)).
- The dipole term produces an additional field that, at the location of R, points toward the sphere (i.e. opposite to the monopole field).
Hence the two contributions partially cancel.
Step 3 – Write the field at R for diagram (c) (outside‑sphere approximation)
For points outside a uniformly charged sphere the field can be expanded in a multipole series. Keeping only the monopole and dipole terms (the next terms are (\propto 1/R^{4}) and are much smaller) gives
[ \mathbf E_{(c)}(R)= \underbrace{\frac{k\,2q}{r^{2}}}{\text{monopole}} \; \hat{\mathbf r} \;+\; \underbrace{\frac{1}{4\pi\varepsilon{0}}\, \frac{3(\mathbf p!\cdot!\hat{\mathbf r})\hat{\mathbf r}-\mathbf p}{r^{3}}}_{\text{dipole}} . ]
Because the induced dipole (\mathbf p) points away from the two +q charges, the scalar product (\mathbf p!\cdot!\hat{\mathbf r}) is negative at point R. The dipole term therefore points opposite to (\hat{\mathbf r}) and reduces the magnitude of the total field.
In symbols, let the dipole contribution at R be (-\,\Delta E) with (\Delta E>0). Then
[ E_{(c)} = \frac{k\,2q}{r^{2}} - \Delta E . ]
Since (\Delta E) is non‑zero (the external +q’s certainly polarise the sphere), we have
[ \boxed{E_{(c)} \;<\; E_{(b)} } . ]
Step 4 – Physical interpretation (intro‑physics language)
- The two +q charges try to push the sphere’s positive charge to the far side.
- Because some of the +2q charge is now farther from point R than before, the net repulsive field that reaches R is weaker.
- The effect is exactly the same as saying: “the sphere looks like a point charge +2q plus a little dipole that pulls the field back toward the sphere.”
Thus the magnitude of the electric field at R is smaller when the point charge is replaced by a conducting sphere carrying the same total charge.
3. Final answer
[ \boxed{\displaystyle \left| \mathbf E_{\text{(c)}}(R) \right| \; < \; \left| \mathbf E_{\text{(b)}}(R) \right| } ]
The field at R in the configuration with the conducting sphere is weaker than the field in the configuration with the isolated +2q point charge.
4. Common mistakes
| Mistake | Why it’s wrong | How to avoid it |
|---|---|---|
| Assuming the sphere’s charge stays uniformly spread | Conductors redistribute charge in response to external fields; the +q charges will polarise the sphere. | Remember that a conductor’s surface charge is not fixed when other charges are nearby. |
| Treating the sphere as a point charge and ignoring induced dipole | The monopole term is correct, but the dipole term (first‑order correction) changes the field direction at R, making it smaller. | Use the multipole idea: “point‑charge field + correction due to induced dipole.” |
| Thinking the fields from the two +q’s do not cancel | Because R is exactly halfway between them, the horizontal components cancel perfectly; only the vertical components (which are zero by symmetry) could remain. | Write the vector expression for the two +q fields and show they sum to zero. |
| Confusing “farther away” with “larger distance from the centre” | The relevant distance is the effective distance of the charge that actually contributes to the field at R. Polarisation moves charge to the far side, increasing that effective distance. | Visualise the induced charge as a little “cloud” shifted away from R. |
| Using the image‑charge method without justification | Introductory students may not know the image‑charge technique, leading to an over‑complicated answer. | Stay with the qualitative monopole‑plus‑dipole picture; a quantitative image‑charge calculation is not required. |
Keeping these points in mind will help students answer the question correctly using only the tools typically taught in a second‑semester introductory physics course.
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