Does the capacitance of a spherical capacitor blow up when the outer sphere is very far away?

No. As R2 approaches infinity, the formula reduces to C = 4πε₀R1, which is the capacitance of an isolated sphere of radius R1. A very distant outer sphere is effectively at infinity.

Does a dielectric-filled capacitor always break down before an air one?

No. Air breaks down at about 3 million volts per meter, while solid dielectrics tolerate 10 to 100 times higher fields. Dielectric-filled capacitors typically handle more voltage before breakdown.

Is the dielectric constant of a material fixed?

Approximately, at low frequencies and moderate fields. At high frequencies, K depends on frequency because molecular dipoles cannot follow the field fast enough. At very high fields, K can also drop.

Which capacitor geometry appears most in JEE and NEET?

Parallel plate capacitor problems dominate in both, especially with dielectric slabs and metal sheets. Spherical capacitors appear more in JEE than NEET. Cylindrical capacitors are relatively rare in NEET.

Capacitance For Geometrical Figures With Dielectric

Q: What is the maximum capacitance you can get from a given parallel plate area and separation?

In the limit K approaching infinity (a perfect conductor filling the gap), capacitance diverges but this is unphysical - the plates would short. Real dielectrics have K up to a few hundred, limited by dielectric breakdown.

Q: When a dielectric is pulled into a capacitor (battery disconnected), does energy increase or decrease?

It decreases. With Q fixed, C increases and U = Q squared over 2C falls. The dielectric is pulled in spontaneously because this energy decrease appears as work done by the electrical force on the dielectric.

PhysicsCapacitorsFor JEE aspirants

The capacitance of a capacitor depends on the geometry of its conductors and the dielectric material between them. Different shapes - parallel plate, spherical, cylindrical - have different capacitance formulas, and inserting a dielectric of constant $K$ always increases capacitance by exactly a factor $K$ . This topic is central to JEE and NEET Physics questions on parallel plate capacitors with dielectric slabs, metal sheets between plates, and spherical or cylindrical geometries.

Key Formulas - Quick Reference

Parallel plate (air): $C = ε_{0} A / d$ ; with dielectric $K$ : $C = K ε_{0} A / d$
Dielectric slabs stacked between plates: $C = ε_{0} A / \sum (t_{i} / K_{i})$
Dielectric regions side by side: $C = (ε_{0} / d) \sum K_{i} A_{i}$
Metal sheet of thickness $t$ between plates: $C = ε_{0} A / (d - t)$
Induced surface charge density: $σ_{i} = σ (1 - 1/ K)$
Spherical (outer earthed): $C = 4 π ε_{0} R_{1} R_{2} / (R_{2} - R_{1})$
Cylindrical (length $ℓ$ ): $C = 2 π ε_{0} K ℓ / ln (R_{2} / R_{1})$
Two connected spheres far apart: $C = 4 π ε_{0} (R_{1} + R_{2})$

1. Dielectrics: How They Work

A dielectric is an insulating material whose atoms or molecules have polar character (either intrinsic or induced by an external field). Common examples: mica, ceramic, paper, glass, distilled water.

Polar dielectrics

These have permanent molecular dipoles that are randomly oriented in the absence of a field. When an external field $E_{0}$ is applied, the dipoles align parallel to it, producing an induced field $E_{i}$ inside the dielectric opposite to $E_{0}$ .

The net field inside the dielectric is:

E = E_{0} - E_{i}, E = \frac{E _{0}}{K}

where

K

is the dielectric constant of the material (always

K \geq 1

;

K = 1

for vacuum).

Induced surface charge

When a dielectric slab fills the gap of a capacitor, induced surface charges appear on its two faces. If $σ$ is the free surface charge density on the capacitor plates, the induced charge density on the dielectric is:

σ_{i} = σ (1 - \frac{1}{K})

These induced charges partially cancel the field from the free charges, leaving a net field reduced by factor $K$ .

2. Parallel Plate Capacitor

Figure: Parallel plate capacitor. Uniform field E fills the gap between plates of area A separated by distance d.

For two parallel conducting plates each of area $A$ separated by distance $d$ with air (or vacuum) between them:

C = \frac{ε _{0} A}{d} (air/vacuum between plates)

If a dielectric of constant $K$ completely fills the gap:

Figure: Parallel plate capacitor with dielectric slab. Capacitance rises by factor K.

C = \frac{K ε _{0} A}{d} and \frac{C _{medium}}{C _{air}} = K

Multiple dielectric slabs stacked between plates

Figure: Stacked dielectric layers between capacitor plates act like capacitors in series.

When several dielectrics of constants $K_{1}, K_{2}, K_{3}, \dots$ and thicknesses $t_{1}, t_{2}, t_{3}, \dots$ (with $t_{1} + t_{2} + t_{3} + \dots = d$ ) are stacked parallel to the plates, the combination acts like capacitors in series:

C = \frac{ε _{0} A}{\frac{t _{1}}{K _{1}} + \frac{t _{2}}{K _{2}} + \frac{t _{3}}{K _{3}} + \dots}

Dielectric regions side by side

Figure: Side-by-side dielectric regions in a parallel plate capacitor act like capacitors in parallel.

When the region between plates is split laterally into patches of areas $A_{1}, A_{2}, A_{3}, \dots$ each with its own dielectric constant $K_{1}, K_{2}, K_{3}, \dots$ (thickness of each equal to full gap $d$ ), the arrangement is equivalent to capacitors in parallel:

C = \frac{ε _{0}}{d} (K_{1} A_{1} + K_{2} A_{2} + K_{3} A_{3} + \dots)

Metal sheet inserted between plates

Figure: Metal sheet of thickness t between capacitor plates. Effective gap becomes (d - t).

If a conducting metal sheet of thickness $t$ is placed anywhere between the plates (not touching either), the metal has $K \to \infty$ effectively. The field inside the metal is zero, so the effective gap becomes $d - t$ :

C = \frac{ε _{0} A}{d - t}

Key observations:

The position of the metal sheet does not affect the capacitance, only its thickness $t$ .
If the sheet is a thin foil ( $t \to 0$ ), the capacitance is unchanged.
If $t = d /2$ , the capacitance doubles: $C = 2 ε_{0} A / d = 2 C_{0}$ .

3. Effect of Battery Connection on a Parallel Plate Capacitor

Whether the battery stays connected or is disconnected changes which quantity is held constant. This distinction is a frequent JEE/NEET trap.

Case A: Battery remains connected (V is constant)

Action	PD (V)	Capacitance (C)	Charge (Q)	Electric field (E)	Energy stored
Reduce plate separation	Same	Increases	Increases	Increases	Increases
Insert dielectric	Same	Increases	Increases	No change	Increases

Case B: Battery disconnected (Q is constant)

Action	Charge (Q)	Capacitance (C)	PD (V)	Electric field (E)	Energy stored
Reduce plate separation	Same	Increases	Decreases	No change	Decreases
Insert dielectric	Same	Increases	Decreases	Decreases	Decreases

Default assumption: if the problem does not specify, assume the battery is disconnected and charge

Q

is constant. Also note: with the battery connected and a dielectric inserted, half the extra energy supplied by the battery is used to pull the dielectric in, and the other half is stored as extra electric potential energy.

4. Spherical Capacitor

Figure: Spherical capacitor. Two concentric shells of radii R₁ and R₂.

Two concentric conducting spheres of radii $R_{1}$ (inner) and $R_{2}$ (outer) form a spherical capacitor.

When the outer sphere is earthed

Figure: Spherical capacitor with outer sphere earthed. Inner has +Q, outer inner-surface has -Q.

C = 4 π ε_{0} \cdot \frac{R _{1} R _{2}}{R _{2} - R _{1}}

With a dielectric of constant $K$ between the spheres, multiply by $K$ .

When the inner sphere is earthed

Figure: Spherical capacitor with inner sphere earthed. Charge q on inner, (Q-q) on outer.

This case behaves as two spherical capacitors in parallel: one between the two spheres, and one between the outer sphere and infinity:

C = \frac{4 π ε _{0} K _{1} R _{1} R _{2}}{R _{2} - R _{1}} + 4 π ε_{0} K_{2} R_{2}

If both media are the same ( $K_{1} = K_{2} = K$ ), this simplifies to:

C = \frac{4 π ε _{0} K R _{2}^{2}}{R _{2} - R _{1}}

Charge distribution when only outer sphere is charged

If charge $Q$ is placed on the outer sphere and the inner sphere is earthed (potential zero), the charge redistributes. Let $q$ be the charge induced on the inner sphere. Setting the inner potential to zero:

q = - \frac{R _{1}}{R _{2}} Q (induced on inner)

Charge on inner surface of outer sphere = \frac{R _{1}}{R _{2}} Q

Charge on outer surface of outer sphere = (1 - \frac{R _{1}}{R _{2}}) Q = \frac{R _{2} - R _{1}}{R _{2}} Q

5. Cylindrical Capacitor

Figure: Cylindrical (coaxial) capacitor of length l. Used in coaxial cables and cylindrical geometries.

Two coaxial cylinders of inner radius $R_{1}$ , outer radius $R_{2}$ and length $ℓ$ (with $ℓ ≫ R_{2}$ so end effects are ignored):

C = \frac{2 π ε _{0} K ℓ}{ln ( R _{2} / R _{1} )} = \frac{2 π ε _{0} K ℓ}{2.303 lo g _{10} ( R _{2} / R _{1} )}

This geometry is common in coaxial cables: signal-carrying inner conductor and grounded outer sheath, separated by an insulating dielectric.

6. Connected Spheres

Two touching spheres of the same radius:

C = 4 π ε_{0} R

Two spheres of radii $R_{1}$ and $R_{2}$ connected by a long wire (far apart):

The two act as parallel capacitors (each connected to a common potential, with the wire acting as the shared conductor):

C = C_{1} + C_{2} = 4 π ε_{0} (R_{1} + R_{2})

Solved Examples

Solved Example 1

A parallel plate capacitor of capacitance

C

has a metal sheet inserted between the plates parallel to them. The thickness of the sheet is half the plate separation. Find the new capacitance.

Solution:

Original capacitance: $C = ε_{0} A / d$ .

With a metal sheet of thickness $t = d /2$ , effective gap becomes $d - t = d - d /2 = d /2$ :

C^{'} = \frac{ε _{0} A}{d - t} = \frac{ε _{0} A}{d /2} = \frac{2 ε _{0} A}{d} = 2 C

Answer: $C^{'} = 2 C$ .

Solved Example 2

The capacitance of an air-filled parallel plate capacitor is

20 pF

. The plate separation is then doubled, and the gap is filled with wax. If the new capacitance is

40 pF

, what is the dielectric constant of wax?

Solution:

Initial capacitance: $C_{0} = ε_{0} A / d = 20 pF$ .

With separation doubled ( $d \to 2 d$ ) and dielectric $K$ filling the gap:

C^{'} = \frac{K ε _{0} A}{2 d} = \frac{K}{2} \cdot C_{0}

Given $C^{'} = 40 pF$ :

40 = \frac{K}{2} \times 20 \Rightarrow K = 4

Note: The source doc gave the initial value as

20 F

and final as

40 \times 1 0^{12} F

(nonsensical values for real capacitors), and the intended answer

K = 8

relies on those specific mismatched units. Working with the sensible

pF

interpretation gives

K = 4

. For JEE, always sanity-check units.

Solved Example 3

A parallel plate air capacitor has plate area

A = 100 cm^{2}

and separation

d = 5 mm

. A potential difference of

300 V

is applied via a battery, which is then disconnected. The space between the plates is then filled with ebonite (

K = 2.6

). Find (a) the initial and final capacitance, (b) the new potential difference, (c) the surface charge density.

Solution:

Convert: $A = 100 \times 1 0^{- 4} m^{2} = 1 0^{- 2} m^{2}$ ; $d = 5 \times 1 0^{- 3} m$ .

(a) Capacitances:

C_{0} = \frac{ε _{0} A}{d} = \frac{8.85 \times 1 0 ^{- 12} \times 1 0 ^{- 2}}{5 \times 1 0 ^{- 3}} = 1.77 \times 1 0^{- 11} F

C = K C_{0} = 2.6 \times 1.77 \times 1 0^{- 11} = 4.6 \times 1 0^{- 11} F

(b) New potential difference. Since the battery is disconnected, charge is constant:

Q = C_{0} V_{0} = 1.77 \times 1 0^{- 11} \times 300 = 5.31 \times 1 0^{- 9} C

V = \frac{Q}{C} = \frac{V _{0}}{K} = \frac{300}{2.6} = 115 V

(c) Surface charge density (unchanged since Q and A are constant):

σ = \frac{Q}{A} = \frac{5.31 \times 1 0 ^{- 9}}{1 0 ^{- 2}} = 5.31 \times 1 0^{- 7} C/m^{2}

Solved Example 4

A spherical condenser has

10 cm

and

12 cm

as the radii of the inner and outer spheres. The space between is filled with a dielectric of constant

3

. Find the capacitance when the outer sphere is earthed.

Solution:

C = \frac{4 π ε _{0} K R _{1} R _{2}}{R _{2} - R _{1}} = \frac{K R _{1} R _{2}}{k ( R _{2} - R _{1} )} where k = 9 \times 1 0^{9}

Plugging in $R_{1} = 0.10 m$ , $R_{2} = 0.12 m$ , $K = 3$ :

C = \frac{3 \times 0.10 \times 0.12}{9 \times 1 0 ^{9} \times 0.02} = \frac{0.036}{1.8 \times 1 0 ^{8}} = 2 \times 1 0^{- 10} F

Answer: $C = 2 \times 1 0^{- 10} F = 200 pF$ .

Solved Example 5

A capacitor stores

10 μ C

of charge when connected across a battery. When the gap between the plates is filled with a dielectric (battery still connected), an additional

20 μ C

flows through the battery. Find the dielectric constant.

Solution:

Battery remains connected, so voltage $V$ is constant.

Without dielectric: $Q_{1} = C V = 10 μ C$ .

With dielectric: total charge is now $Q_{1} + 20 = 30 μ C$ . And this equals $K C V$ :

Q_{2} = K C V = 30 μ C \Rightarrow \frac{Q _{2}}{Q _{1}} = K = 3

Answer: $K = 3$ .

Solved Example 6

Four large metal plates are located parallel to each other, close together. The extreme (outermost) plates are connected by a wire, while a potential difference

V

is applied between the two inner plates. Find the electric field between the neighboring plates.

Solution:

Label the plates 1, 2, 3, 4 from left to right. Plates 1 and 4 are shorted together via the external wire; the potential $V$ is applied between plates 2 and 3.

Let $V^{'}$ be the potential difference between plates 1-2 (which equals that between 3-4 by symmetry, since 1 and 4 are at the same potential and the geometry is symmetric).

Applying Kirchhoff's voltage rule around the loop $1 \to 2 \to 3 \to 4 \to 1$ :

- V^{'} + V + V^{'} - 0 = 0 \Rightarrow V^{'} = 0 ?

Reconsidering with signs: going from plate 1 to 2 we rise by $V^{'}$ ; from 2 to 3 we drop by $V$ ; from 3 to 4 we drop by $V^{'}$ ; from 4 back to 1 (via the wire) we do zero. Sum $= V^{'} - V - V^{'} = - V$ . For a valid loop this must be zero, which is impossible unless $V = 0$ . The paradox resolves by noting that plate 1 (and plate 4) are floating charges: they redistribute such that the total flux inside the sandwich is zero.

Careful accounting: field between plates 2 and 3 is $E_{23} = V / d$ (basic parallel plate). Fields in the two outer gaps have equal magnitude $E^{'} = V^{'} / d$ but their directions oppose the inner field. Since the enclosed net charge on plates 1 and 4 must sum to zero, the outer fields carry half the inner-plate charge in each direction, giving $E^{'} = V / (2 d)$ .

Answer: Field between the inner pair of plates is $V / d$ ; field in each outer gap is $V / (2 d)$ .

Common Mistakes to Avoid

Watch out

Assuming battery connection state without checking. Read the problem: "battery connected" means $V$ constant; "battery disconnected" means $Q$ constant. The two cases give opposite answers for how energy changes when a dielectric is inserted.
Treating stacked dielectric slabs as parallel instead of series. Slabs stacked between the plates (perpendicular to the field) act like series capacitors ( $1/ C$ formula). Slabs side by side (parallel to the field) act like parallel capacitors ( $C_{i}$ sum).
Forgetting that metal sheet position does not matter. Only the thickness $t$ affects the capacitance in $C = ε_{0} A / (d - t)$ .
Using $C = ε_{0} A / d$ for the spherical capacitor. The parallel plate formula is a specific geometry; spherical geometry uses $C = 4 π ε_{0} R_{1} R_{2} / (R_{2} - R_{1})$ .
Missing the induced surface charge relation. When a dielectric is inserted, the free surface charge on the plates may or may not change, but the induced charge on the dielectric is always $σ_{i} = σ (1 - 1/ K)$ .

Frequently Asked Questions

Q1. Why does adding a dielectric always increase capacitance?

The dielectric molecules polarize under the field, creating induced charges on the dielectric surfaces that partly cancel the field between the plates. For the same charge on the plates, the potential difference is reduced by factor $K$ , so $C = Q / V$ increases by factor $K$ .

Q2. What is the maximum capacitance you can get from a given parallel plate area and separation?

In the limit $K \to \infty$ (a perfect conductor filling the gap), the capacitance formula $C = K ε_{0} A / d$ diverges, but this is unphysical - the plates would short. Real dielectrics have $K$ up to a few hundred (barium titanate reaches $K \approx 1200$ ), limited by dielectric breakdown at strong fields.

Q3. Why does the position of a metal sheet inserted between capacitor plates not matter?

The electric field is zero inside the metal, so the metal effectively removes a slab of thickness $t$ from the effective gap. Whether the metal sits near the top plate, the middle, or the bottom, the total "field-carrying" gap remains $d - t$ , giving the same capacitance $ε_{0} A / (d - t)$ .

Q4. Does the capacitance of a spherical capacitor with outer radius $R_{2}$ blow up when the outer sphere is very far away?

Yes and no. As $R_{2} \to \infty$ , $R_{1} R_{2} / (R_{2} - R_{1}) \to R_{1}$ , so $C \to 4 π ε_{0} R_{1}$ . This equals the capacitance of an isolated sphere of radius $R_{1}$ - which makes physical sense, since a very distant outer sphere is effectively "at infinity."

Q5. Why do coaxial cables use a cylindrical capacitor design?

The cylindrical geometry naturally confines the electric field between the inner conductor and the outer grounded sheath, preventing signal leakage and shielding from external interference. Its capacitance per unit length, $2 π ε_{0} K / ln (R_{2} / R_{1})$ , is easily controlled by choosing dielectric material and geometry.

Q6. When two capacitors are physically identical except one has a dielectric, does the one without a dielectric always break down first?

Not necessarily. Dielectric breakdown occurs at a characteristic field strength (breakdown field). Air breaks down at about $3 \times 1 0^{6} V/m$ , while many solid dielectrics tolerate $1 0^{7}$ - $1 0^{8} V/m$ . So a dielectric-filled capacitor typically handles more voltage before breakdown, even though its stored energy per unit volume is higher.

Q7. Is the dielectric constant of a material a fixed number?

Approximately, at low frequencies and moderate fields. At high frequencies (radio, microwave), $K$ depends on frequency because molecular dipoles cannot follow the field fast enough. At very high fields, $K$ can also drop and the material approaches breakdown.

Q8. For JEE and NEET, which geometry appears most often in problems?

Parallel plate capacitor problems dominate in both exams, especially with dielectric slabs (stacked or side by side) and metal sheets inserted. Spherical capacitor problems appear less often but are common in JEE Main and Advanced. Cylindrical capacitors are relatively rare in NEET but appear in JEE occasionally, usually via coaxial cable questions.

Q9. When a dielectric is being pulled into a parallel plate capacitor (battery disconnected), does the total energy increase or decrease?

It decreases. With $Q$ fixed, $C$ increases and $U = Q^{2} / (2 C)$ falls. The dielectric is pulled in spontaneously because this energy decrease appears as work done by the electrical force on the dielectric. This is the physical origin of the "force on a dielectric" formula.

Previous year questions on Capacitance For Geometrical Figures With Dielectric

16 questions from past papers, each with a step-by-step solution.

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1. Dielectrics: How They Work

Polar dielectrics

Induced surface charge

2. Parallel Plate Capacitor

Multiple dielectric slabs stacked between plates

Dielectric regions side by side

Metal sheet inserted between plates

3. Effect of Battery Connection on a Parallel Plate Capacitor

Case A: Battery remains connected (V is constant)

Case B: Battery disconnected (Q is constant)

4. Spherical Capacitor

When the outer sphere is earthed

When the inner sphere is earthed

Charge distribution when only outer sphere is charged

5. Cylindrical Capacitor

6. Connected Spheres

Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why does adding a dielectric always increase capacitance?

Q2. What is the maximum capacitance you can get from a given parallel plate area and separation?

Q3. Why does the position of a metal sheet inserted between capacitor plates not matter?

Q4. Does the capacitance of a spherical capacitor with outer radius R2​ blow up when the outer sphere is very far away?

Q5. Why do coaxial cables use a cylindrical capacitor design?

Q6. When two capacitors are physically identical except one has a dielectric, does the one without a dielectric always break down first?

Q7. Is the dielectric constant of a material a fixed number?

Q8. For JEE and NEET, which geometry appears most often in problems?

Q9. When a dielectric is being pulled into a parallel plate capacitor (battery disconnected), does the total energy increase or decrease?

Previous year questions on Capacitance For Geometrical Figures With Dielectric

Ready to master Capacitors?

Q4. Does the capacitance of a spherical capacitor with outer radius $R_{2}$ blow up when the outer sphere is very far away?