Physics and Mathematics
Spherical Capacitor
1. Concept Overview
A spherical capacitor is a capacitor formed by two concentric spherical conductors—
- an inner sphere of radius [a], and
- an outer spherical shell of radius [b], where [b > a].
The region between them is filled with a dielectric (usually air or vacuum).
When a charge [+Q] is placed on the inner sphere, the outer sphere gets an induced charge [-Q].
The electric field exists only in the region between the spheres, not inside the inner conductor or outside the outer conductor.
A spherical capacitor is important because:
- It provides a real 3D geometry for studying capacitors.
- The field is symmetrical and purely radial, making it easy to apply Gauss’s Law.
- Its capacitance is independent of the amount of charge.
2. Mathematical Derivation
Electric Field Between the Spheres
Using Gauss’s law, for [a < r < b]:
[E(r)] [= \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}]
Potential Difference Between the Plates
Potential difference:
[V = \int_{a}^{b} E(r) ,] [dr= \int_{a}^{b} \dfrac{1}{4\pi\varepsilon_0} \dfrac{Q}{r^2} dr]
[V] [= \dfrac{Q}{4\pi\varepsilon_0} \left( \dfrac{1}{a} – \dfrac{1}{b} \right)]
Capacitance
[C = \dfrac{Q}{V}] [= \dfrac{4\pi\varepsilon_0 ab}{b – a}]
If a dielectric of permittivity [\varepsilon = \varepsilon_r \varepsilon_0] is present:
[C] [= \dfrac{4\pi\varepsilon ab}{b – a}]
3. Dimensions and Units
- Dimension of capacitance:
[M^{-1} L^{-2} T^{4} A^{2}] - SI Unit: Farad (F)
4. Key Features
- Uses spherical symmetry → electric field depends only on distance [r].
- Field exists only between [a < r < b].
- Outer region ([r > b]) has zero field (net charge = 0).
- Capacitance increases when:
- The gap [b – a] decreases
- The dielectric constant [\varepsilon_r] increases
- Radii [a] and [b] are larger
- No fringing effect (ideal geometry).
5. Important Formulas to Remember
| Quantity | Formula |
|---|---|
| Electric field | [E(r)] [= \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}] |
| Potential difference | [V] [= \dfrac{Q}{4\pi\varepsilon_0}\left( \dfrac{1}{a} – \dfrac{1}{b} \right)] |
| Capacitance | [C] [= \dfrac{4\pi\varepsilon_0 ab}{b – a}] |
| With dielectric | [C] [= \dfrac{4\pi\varepsilon ab}{b – a}] |
6. Conceptual Questions with Solutions
1. Why does the electric field exist only between the spheres?
Inside the inner sphere and outside the outer sphere, the net enclosed charge is zero. By Gauss’s law, electric field is zero in these regions.
2. Why is the electric field radial?
Because the geometry is spherically symmetric. Any non-radial component would violate symmetry.
3. Does the outer sphere contribute to the electric field?
No, because for **[a < r < b]**, only the charge on the inner sphere is enclosed by the Gaussian surface.
4. Why is potential found by integrating E?
Because **[V = -\int \vec{E} \cdot d\vec{r}]**, based on the work done in moving a unit positive charge.
5. Why is capacitance not dependent on charge Q?
Capacitance depends only on geometry and dielectric medium, not on the stored charge.
6. What happens to C if b → ∞?
You get a single isolated sphere. [ C = 4\pi\varepsilon_0 a ]
7. Why does increasing dielectric constant increase capacitance?
Because the dielectric reduces electric field and potential difference, allowing more charge to be stored.
8. Does the sign of charge affect capacitance?
No, capacitance is always positive and charge-independent.
9. Why is potential difference inversely proportional to radii?
Because V involves terms **[1/a]** and **[1/b]**, derived from integrating **[E \propto 1/r^2]**.
10. Does changing both radii but keeping b − a same change C?
Yes, because C depends on product **[ab]**, not just the gap.
11. Is the electric field continuous at r = a?
No, it changes abruptly from zero (inside inner conductor) to a finite value just outside.
12. Why is no charge present inside the dielectric?
Because charges reside only on the surface of conductors in electrostatic equilibrium.
13. Why does E fall as 1/r²?
Due to spherical symmetry and Gauss’s law.
14. Why is potential positive for the inner sphere?
A positive charge produces positive potential everywhere around it relative to infinity.
15. What happens if the spheres touch?
The capacitor is destroyed—no isolation → no charge separation → zero capacitance.
7. FAQ / Common Misconceptions
1. Is the electric field inside the inner sphere non-zero?
No. The inner sphere is a conductor → field inside is always zero.
2. Does the outer sphere feel the same field everywhere?
No, field varies as **[1/r^2]** inside the dielectric region.
3. More charge means more capacitance?
Incorrect. Capacitance is purely geometric.
4. Potential is zero outside the capacitor?
No. Only electric field is zero; potential is constant, not zero.
5. Capacitance decreases if dielectric constant increases?
No. It increases.
6. Larger gap always means larger capacitance?
No, larger gap means **smaller** capacitance.
7. Does the outer sphere store charge uniformly?
Yes, because it is also spherical and metallic.
8. Can charge reside inside the gap?
No, it only stays on the conductor surfaces.
9. Is potential difference the same as electric field?
No. E is a local quantity; V is a global quantity depending on two points.
10. Does increasing radius a always increase capacitance?
Not always—it also depends on b and the ratio **[ab/(b – a)]**.
8. Practice Questions (With Step-by-Step Solutions)
Q1. A spherical capacitor has radii [a = 5 cm] and [b = 10 cm]. Calculate its capacitance.
Solution:
Convert to meters:
[a = 0.05 m], [b = 0.10 m]
[
C = \dfrac{4\pi\varepsilon_0 ab}{b – a}
]
[C] [= \dfrac{4\pi(8.85\times 10^{-12})(0.05)(0.10)}{0.05}]
[C] [= 4\pi \times 8.85\times 10^{-12}\times 0.1]
[C] [\approx 1.11\times 10^{-11} \text{ F}]
Q2. The potential difference between the spheres of a spherical capacitor is 200 V. If the inner sphere carries +2 μC, find C.
[
C = \dfrac{Q}{V}
]
[
C = \dfrac{2\times 10^{-6}}{200}
]
[
C = 1\times 10^{-8} \text{ F}
]
Q3. Find V between the spheres if [a = 5 cm], [b = 15 cm], and [Q = 4 μC].
[V] [= \dfrac{Q}{4\pi\varepsilon_0} \left( \dfrac{1}{a} – \dfrac{1}{b} \right)]
Convert radii to meters:
[a = 0.05], [b = 0.15]
[V] [= \dfrac{4\times 10^{-6}}{4\pi(8.85\times 10^{-12})}] [\times\left( \dfrac{1}{0.05} – \dfrac{1}{0.15} \right)]
Compute the bracket:
[\dfrac{1}{0.05}] [= 20,] [\quad] [\dfrac{1}{0.15} = 6.67]
[
20 – 6.67 = 13.33
]
Now:
[V] [\approx \dfrac{4\times 10^{-6}}{1.11\times 10^{-10}} \times 13.33]
[
V \approx 4.8\times 10^{5} \text{ V}
]
Q4. If the dielectric constant between the spheres is 4, how does capacitance change?
[C_{\text{new}}] [= \varepsilon_r C = 4C]
Capacitance becomes 4 times the original value.
Q5. A spherical capacitor has [a = 10 cm], [b = 20 cm], filled with dielectric [\varepsilon_r = 5]. Find its capacitance.
[
C = \dfrac{4\pi\varepsilon ab}{b – a}
]
[
= \dfrac{4\pi(5\varepsilon_0)(0.1)(0.2)}{0.1}
]
[
= 4\pi\times 5\varepsilon_0\times 0.2
]
[
= 4\pi\times (8.85\times 10^{-12})\times 1
]
[
C \approx 1.11\times 10^{-10} \text{ F}
]
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