Capacitor flash cards

1. 1.What is a capacitor?

   A device that stores electric charge and electrical energy, consisting of two conductors separated by an insulator (dielectric). It stores equal and opposite charges $+Q$ and $-Q$ on its two plates.

   Hint: Two conductors + insulator.

2. 2.Define capacitance of a capacitor.

   The ratio of charge on either plate to the potential difference between the plates: $C = \dfrac{Q}{V}$. It measures a conductor's ability to store charge per unit potential.

   Hint: $C=Q/V$

3. 3.What is the SI unit of capacitance? Define it.

   The farad (F). $1\,\text{F} = 1\,\text{C/V}$ — i.e. 1 farad is the capacitance when 1 coulomb of charge raises the potential by 1 volt.

   Hint: Coulomb per volt.

4. 4.Why is the farad a very large unit? What practical units are used?

   1 F requires enormous charge for 1 V, so practical capacitors use sub-multiples: $1\,\mu\text{F}=10^{-6}\,\text{F}$, $1\,\text{nF}=10^{-9}\,\text{F}$, $1\,\text{pF}=10^{-12}\,\text{F}$.

   Hint: microfarad, nanofarad, picofarad.

5. 5.On what factors does the capacitance of a capacitor depend?

   On the geometry (size, shape, separation of conductors) and the medium (dielectric) between them. It does not depend on the charge or potential given to it.

   Hint: Geometry and medium only.

6. 6.Write the formula for the capacitance of an isolated spherical conductor of radius $R$.

   $C = 4\pi\varepsilon_0 R$. A single sphere acts like a capacitor with its other plate at infinity.

   Hint: $C\propto R$.

7. 7.What is the capacitance of an isolated sphere of radius 1 m in vacuum (order of magnitude)?

   $C = 4\pi\varepsilon_0 R = \dfrac{1}{9\times10^{9}}\times 1 \approx 1.1\times10^{-10}\,\text{F} \approx 111\,\text{pF}$. This shows why the farad is huge — even Earth's capacitance is only about $700\,\mu\text{F}$.

   Hint: $1/(9\times10^9)$.

8. 8.Derive the capacitance of a parallel plate capacitor (vacuum).

   Field between plates $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 A}$; potential difference $V=Ed=\dfrac{Qd}{\varepsilon_0 A}$. So $C=\dfrac{Q}{V}=\dfrac{\varepsilon_0 A}{d}$.

   Hint: $V=Ed$, then $C=Q/V$.

9. 9.State the capacitance formula for a parallel plate capacitor and identify each symbol.

   $C=\dfrac{\varepsilon_0 A}{d}$, where $A$ = plate area, $d$ = separation, $\varepsilon_0=8.85\times10^{-12}\,\text{F/m}$ is the permittivity of free space.

   Hint: Area over distance.

10. 10.How does the capacitance of a parallel plate capacitor change if plate area is doubled and separation is halved?

    $C=\dfrac{\varepsilon_0 A}{d}$. Doubling $A$ doubles $C$; halving $d$ doubles $C$ again. Net effect: capacitance becomes 4 times the original.

    Hint: $C\propto A$, $C\propto 1/d$.

11. 11.What is the value of the permittivity of free space $\varepsilon_0$?

    $\varepsilon_0 = 8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$ (or F/m). It relates to Coulomb's constant by $\dfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\text{Nm}^2\text{C}^{-2}$.

    Hint: $8.85\times10^{-12}$.

12. 12.What is the electric field between the plates of a parallel plate capacitor?

    $E=\dfrac{\sigma}{\varepsilon_0}=\dfrac{Q}{\varepsilon_0 A}$. It is uniform and directed from the positive to the negative plate (ignoring edge effects).

    Hint: $\sigma/\varepsilon_0$.

13. 13.What is the field just outside the plates of an ideal parallel plate capacitor?

    Zero (ideally). The fields due to the two oppositely charged plates cancel outside and add up (to $\sigma/\varepsilon_0$) only in the region between them.

    Hint: Fields cancel outside.

14. 14.Why does each plate carry surface charge density that gives field $\sigma/2\varepsilon_0$, yet the field between plates is $\sigma/\varepsilon_0$?

    A single charged sheet produces $\dfrac{\sigma}{2\varepsilon_0}$ on each side. Between the plates the two sheets' fields point the same way and add: $\dfrac{\sigma}{2\varepsilon_0}+\dfrac{\sigma}{2\varepsilon_0}=\dfrac{\sigma}{\varepsilon_0}$.

    Hint: Two sheets add up between.

15. 15.What is a dielectric?

    A non-conducting (insulating) material such as glass, mica, paper or air. When placed in a capacitor it increases capacitance by reducing the net field through polarization.

    Hint: Insulator that polarizes.

16. 16.Define the dielectric constant (relative permittivity) $K$.

    $K=\dfrac{C_\text{medium}}{C_\text{vacuum}}=\dfrac{\varepsilon}{\varepsilon_0}$ — the factor by which capacitance increases when the gap is filled with the dielectric. It is dimensionless and $\geq 1$.

    Hint: Ratio of capacitances.

17. 17.How does inserting a dielectric of constant $K$ change the capacitance of a parallel plate capacitor?

    It multiplies capacitance by $K$: $C=\dfrac{K\varepsilon_0 A}{d}=\dfrac{\varepsilon A}{d}$, where $\varepsilon=K\varepsilon_0$ is the permittivity of the medium.

    Hint: Multiply by $K$.

18. 18.What happens to the electric field inside a dielectric placed in an external field?

    The field is reduced to $E=\dfrac{E_0}{K}$, where $E_0$ is the applied field. Polarization creates an opposing field that partially cancels the applied one.

    Hint: Reduced by factor $K$.

19. 19.What is polarization of a dielectric?

    The alignment of molecular dipoles (or induced dipoles) along the applied field, producing bound surface charges. This sets up an internal field opposing the external one, reducing the net field.

    Hint: Dipoles align, bound charges appear.

20. 20.Distinguish polar and non-polar dielectrics.

    Polar molecules (e.g. $\text{H}_2\text{O}$, HCl) have a permanent dipole moment; they align in a field. Non-polar molecules (e.g. $\text{O}_2$, $\text{H}_2$) have no permanent dipole but acquire an induced one in a field.

    Hint: Permanent vs induced dipole.

21. 21.What is dielectric strength?

    The maximum electric field a dielectric can withstand without breaking down (becoming conducting). For air it is about $3\times10^{6}\,\text{V/m}$. It limits the maximum voltage on a capacitor.

    Hint: Max field before breakdown.

22. 22.State the capacitance of a parallel plate capacitor completely filled with a dielectric slab of constant $K$ and thickness equal to the gap $d$.

    $C=\dfrac{K\varepsilon_0 A}{d}$. The whole field region is dielectric, so capacitance rises $K$-fold.

    Hint: Slab fills whole gap.

23. 23.A dielectric slab of thickness $t$ ($t<d$) and constant $K$ is inserted in a parallel plate capacitor of gap $d$. Write the capacitance.

    $C=\dfrac{\varepsilon_0 A}{d-t+\dfrac{t}{K}}$. The slab effectively reduces the air gap by $t\left(1-\dfrac{1}{K}\right)$.

    Hint: Replace $d$ by $d-t+t/K$.

24. 24.A conducting slab of thickness $t$ ($t<d$) is inserted between the plates. Find the new capacitance.

    $C=\dfrac{\varepsilon_0 A}{d-t}$ (put $K\to\infty$ in the dielectric formula). A conductor is equivalent to a dielectric of infinite $K$; only the air gap $d-t$ matters.

    Hint: Conductor = $K\to\infty$.

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