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Animated Solution for Physics - Electrostatics: A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

Q: What would happen if the aluminium foil had a significant thickness?

If the foil had a non-negligible thickness $t$, the effective distance between the plates would decrease to $(d - t)$. According to the formula $C = \frac{\varepsilon_0 A}{d - t}$, the capacitance would increase.

Q: Can I solve this by treating it as two capacitors in series?

Absolutely! If the foil is at distance $x$ from one plate, you have two capacitors: $C_1 = \frac{\varepsilon_0 A}{x}$ and $C_2 = \frac{\varepsilon_0 A}{d - x}$. Using the series formula $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$, you will find that $C_{eq} = \frac{\varepsilon_0 A}{d}$, which is the same result.

Select Answer:

Visualized Solution

Initial Setup of the Capacitor

Let the initial capacitance of the parallel plate capacitor be $C_{0}$ .
C_0 = \frac{\varepsilon_0 A}{d}
where $A$ is the area of the plates and $d$ is the separation between them.

Introducing the Aluminium Foil

An aluminium foil is introduced between the plates.
Aluminium is a metal, which acts as a perfect conductor.

Induced Charges on the Foil

Equal and opposite charges are induced on the two surfaces of the conducting foil.
However, the foil has a negligible thickness ( $t \to 0$ ).

Capacitance with a Dielectric Slab

The capacitance of a capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is:
C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}

Applying Conditions for Aluminium Foil

For a conducting metal like aluminium, the dielectric constant $K = \infty$ .
The thickness of the foil is negligible, so $t \to 0$ .

Calculating Final Capacitance

Substitute $K = \infty$ and $t = 0$ into the formula:
C = \frac{\varepsilon_0 A}{d - 0 + \frac{0}{\infty}}
C = \frac{\varepsilon_0 A}{d}

Conclusion

C = C_0
The capacitance remains unchanged.

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The Sigma Insight: Capacitance and Capacitors

Solution Diagram

The Illusion of the Aluminium Foil

Why Capacitance Remains Unchanged

Imagine you are working with a standard parallel plate capacitor in a lab. It consists of two metallic plates separated by a distance

d

, with a vacuum or air in between. The initial capacitance of this setup is beautifully simple, given by the formula:

C_{0} = \frac{ε _{0} A}{d}

Now, let's introduce a twist. You take a sheet of aluminium foil and slide it right between the plates. What happens to the capacitance? Does it increase, decrease, or stay exactly the same? To answer this, we need to dive into the physics of conductors and electric fields.

The Physics of the Foil

Aluminium is a metal, which means it is an excellent conductor of electricity. When a conductor is placed in an external electric field (like the one between the capacitor plates), the free electrons inside the metal immediately redistribute themselves. They move until they completely cancel out the electric field inside the conductor.

This redistribution results in equal and opposite charges being induced on the two surfaces of the foil. However, the problem gives us a crucial piece of information: the foil has negligible thickness (

t \to 0

The Mathematical Proof

To see how this affects the capacitance, we can use the general formula for a capacitor partially filled with a dielectric slab of thickness

t

and dielectric constant

K

C = \frac{ε _{0} A}{d - t + \frac{t}{K}}

Here is where the magic happens. Because aluminium is a perfect conductor, its dielectric constant

K

is effectively infinity (

K = \infty

). This means the term

\frac{t}{K}

becomes exactly zero.

Furthermore, because the foil is incredibly thin, its thickness

t

approaches zero. Let's substitute these conditions into our master equation:

C = \frac{ε _{0} A}{d - 0 + 0}

The Final Verdict

Simplifying the expression, we are left with:

C = \frac{ε _{0} A}{d}

This is exactly equal to our initial capacitance,

C_{0}

. The introduction of the infinitely thin conducting foil did absolutely nothing to alter the overall ability of the system to store charge. Therefore, the capacitance remains unchanged. It's a beautiful example of how physical intuition and mathematical rigor align perfectly in electrostatics!

Common Traps & FAQs

Why is the dielectric constant (

K

) of a metal considered to be infinity?

In a perfect conductor, free electrons move to cancel any internal electric field, making the net electric field inside the metal exactly zero. Since

E_{in s i d e} = \frac{E _{e x t er na l}}{K}

, for

E_{in s i d e}

to be zero,

K

must approach infinity.

What would happen if the aluminium foil had a significant thickness?

If the foil had a non-negligible thickness

t

, the effective distance between the plates would decrease to

(d - t)

. According to the formula

C = \frac{ε _{0} A}{d - t}

, the capacitance would increase.

Does the position of the foil between the plates matter?

No, the position of the foil does not matter. The formula for a dielectric or conducting slab depends only on its thickness

t

, not on its exact location between the plates.

Why do we even care about the induced charges on the foil?

The induced charges are what physically cause the electric field inside the foil to become zero. This is the fundamental reason why a metal acts differently than an insulator (dielectric) when placed in a capacitor.

Can I solve this by treating it as two capacitors in series?

Absolutely! If the foil is at distance

x

from one plate, you have two capacitors:

C_{1} = \frac{ε _{0} A}{x}

and

C_{2} = \frac{ε _{0} A}{d - x}

. Using the series formula

\frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{2}}

, you will find that

C_{e q} = \frac{ε _{0} A}{d}

, which is the same result.

Animated Solution for Physics - Electrostatics: A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

Select Answer:

Visualized Solution

Initial Setup of the Capacitor

Introducing the Aluminium Foil

Induced Charges on the Foil

Capacitance with a Dielectric Slab

Applying Conditions for Aluminium Foil

Calculating Final Capacitance

Conclusion

The Sigma Insight: Capacitance and Capacitors

The Illusion of the Aluminium Foil

The Physics of the Foil

The Mathematical Proof

The Final Verdict

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