Capacitance is the charge stored per unit potential difference between the plates of a capacitor.

What is the unit of capacitance , and why are smaller prefixed units commonly used in practice?

Capacitance is measured in farads (F) . In practice, 1 F is a very large unit, so capacitance is often quoted in microfarads (μF), nanofarads (nF) or picofarads (pF).

True or False? In the capacitance equation, the charge Q is the charge stored by the capacitor as a whole.

False. The charge Q is the charge stored on the plates . The capacitor itself does not store net charge.

The greater the capacitance of a capacitor, the greater the .......... stored.

The greater the capacitance of a capacitor, the greater the energy stored.

Define polar molecule .

A polar molecule is a molecule that has a 'positive' end and a 'negative' end (poles).

How are the polar molecules in a dielectric arranged when no charge is applied to the capacitor?

They are aligned in random directions, since there is no electric field between the plates.

How do the polar molecules in a dielectric behave once a charge is applied to the capacitor plates?

An electric field is generated between the plates. The negative ends of the polar molecules are attracted to the positive plate (and vice versa), so all the molecules rotate to align themselves parallel to the electric field .

Define relative permittivity (dielectric constant).

Relative permittivity is the ratio of the permittivity of a material to the permittivity of free space.

True or False? Relative permittivity has units of F m -1 .

False. Relative permittivity has no units , because it is a ratio of two quantities with the same unit.

What effect does the electric field produced by aligned polar molecules in a dielectric have on the field from the capacitor plates?

It opposes the electric field produced by the plates, reducing the overall electric field between them.

For a capacitor holding a fixed charge, how does adding a dielectric with a larger opposing field (i.e. larger permittivity) affect the potential difference and capacitance?

The potential difference between the plates decreases , so the capacitance of the capacitor increases .

In the parallel plate capacitance equation, A refers to the cross-sectional area of .......... of the plates, not both.

In the parallel plate capacitance equation, A refers to the cross-sectional area of one of the plates, not both.

Why does more work need to be done to add further charge to a capacitor's negative plate as it charges?

As the negative plate becomes more negatively charged, the force of repulsion between the electrons already on the plate and the new electrons being added increases , so more work is needed.

Since charge Q is directly proportional to potential difference V , a graph of Q against V for a capacitor is a straight line through the .......... .

Since charge Q is directly proportional to potential difference V , a graph of Q against V for a capacitor is a straight line through the origin .

How can the energy stored in a capacitor be determined from a graph of potential difference against charge?

It is equal to the area under the graph .

A LevelPhysicsAQAFlashcardsFields & Their ConsequencesCapacitance

Capacitance (AQA A Level Physics): Flashcards

Exam code: 7408

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FrontCapacitance
Define capacitance.
FrontCapacitance
State the equation for capacitance and define its symbols.
BackCapacitance
$C = \frac{Q}{V}$
C = capacitance (F)
Q = charge (C)
V = potential difference (V)

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Cards in this collection (23)

Define capacitance.
Capacitance is the charge stored per unit potential difference between the plates of a capacitor.
State the equation for capacitance and define its symbols.
$C = \frac{Q}{V}$
- C = capacitance (F)
- Q = charge (C)
- V = potential difference (V)
What is the unit of capacitance, and why are smaller prefixed units commonly used in practice?
Capacitance is measured in farads (F). In practice, 1 F is a very large unit, so capacitance is often quoted in microfarads (μF), nanofarads (nF) or picofarads (pF).
True or False?
In the capacitance equation, the charge Q is the charge stored by the capacitor as a whole.
False.
The charge Q is the charge stored on the plates. The capacitor itself does not store net charge.
The greater the capacitance of a capacitor, the greater the .......... stored.
The greater the capacitance of a capacitor, the greater the energy stored.
Define polar molecule.
A polar molecule is a molecule that has a 'positive' end and a 'negative' end (poles).
How are the polar molecules in a dielectric arranged when no charge is applied to the capacitor?
They are aligned in random directions, since there is no electric field between the plates.
How do the polar molecules in a dielectric behave once a charge is applied to the capacitor plates?
An electric field is generated between the plates. The negative ends of the polar molecules are attracted to the positive plate (and vice versa), so all the molecules rotate to align themselves parallel to the electric field.
Define relative permittivity (dielectric constant).
Relative permittivity is the ratio of the permittivity of a material to the permittivity of free space.
State the equation for relative permittivity and define its symbols.
$ϵ_{r} = \frac{ϵ}{ϵ_{0}}$
- ε_r = relative permittivity
- ε = permittivity of the material (F m^-1)
- ε₀ = permittivity of free space (F m^-1)
True or False?
Relative permittivity has units of F m^-1.
False.
Relative permittivity has no units, because it is a ratio of two quantities with the same unit.
What effect does the electric field produced by aligned polar molecules in a dielectric have on the field from the capacitor plates?
It opposes the electric field produced by the plates, reducing the overall electric field between them.
For a capacitor holding a fixed charge, how does adding a dielectric with a larger opposing field (i.e. larger permittivity) affect the potential difference and capacitance?
The potential difference between the plates decreases, so the capacitance of the capacitor increases.
State the equation for the capacitance of a parallel plate capacitor with a dielectric, and define its symbols.
$C = \frac{ϵ_{r} ϵ_{0} A}{d}$
- C = capacitance (F)
- A = cross-sectional area of the plates (m²)
- d = separation of the plates (m)
- ε_r = relative permittivity of the dielectric
- ε₀ = permittivity of free space (F m^-1)
In the parallel plate capacitance equation, A refers to the cross-sectional area of .......... of the plates, not both.
In the parallel plate capacitance equation, A refers to the cross-sectional area of one of the plates, not both.
If a parallel plate capacitor has square plates of side length L, what is the cross-sectional area A?
$A = L^{2}$
Why does more work need to be done to add further charge to a capacitor's negative plate as it charges?
As the negative plate becomes more negatively charged, the force of repulsion between the electrons already on the plate and the new electrons being added increases, so more work is needed.
Since charge Q is directly proportional to potential difference V, a graph of Q against V for a capacitor is a straight line through the ...........
Since charge Q is directly proportional to potential difference V, a graph of Q against V for a capacitor is a straight line through the origin.
How can the energy stored in a capacitor be determined from a graph of potential difference against charge?
It is equal to the area under the graph.
State the equation for energy stored by a capacitor in terms of charge Q and potential difference V.
$E = \frac{1}{2} Q V$
State the equation for energy stored by a capacitor in terms of capacitance C and potential difference V.
$E = \frac{1}{2} C V^{2}$
State the equation for energy stored by a capacitor in terms of charge Q and capacitance C.
$E = \frac{Q^{2}}{2 C}$
True or False?
Doubling the potential difference across a capacitor doubles the energy stored in it.
False.
Since $E = \frac{1}{2} C V^{2}$ , energy stored is proportional to V², so doubling V quadruples the energy stored.

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