**4.Capacitance and Capacitors.**

**a.
Parallel Plate Capacitor.**

Consider the diagram below with two sets of parallel plates, labeled A and B, and twice as many charges distributed at time t2 than time t1:

* Fig. 4: Plate pairs with increasing charge at later time** t2*

We see that the p.d. between the plates A and B doubles in the time interval (t2 – t1). Or: V(t1) = 2 V(t2).

Since the Electric field is E = V/d then: E(t2) = 2E(t1). Note that the distance between plates d is kept constant. Experimentally we find the following true:

i)
The
quantity of charge Q ~ V

**ii)
**The
quantity of charge Q ~ 1/d

**iii)
**The
quantity of charge Q ~ A

Where A denotes
the area of the plates. Look now at the surface charge density: s = Q/ A and we know: E e_{o} = s.

**Then
we have: **

E = Q/ e_{o} A
or: Q = E e_{o} A

Taking experimental data for (i)- (iii) together:

**Q
****~**** V A /d **

**But:
definition of capacitance is charge stored per unit potential difference
between the plates or C = Q/V.**

**Therefore:
Q = CV**

This
in turn implies: CV **~**** **V**
**A /d** **

Or:
C **~**** ** ** **A
/d** **and C = k (A/D)

k =
constant = e_{o}

Hence:
C** =** e_{o}** **A/d

We
may also write: C **=** e_{o }e_{r}** **A/d

**Where
**e_{r }** is the
relative permittivity of the medium.**

Consider Fig. 5 below:

** Fig. 5: Parallel plates separated by d with dielectric between**

Here the capacitor
has a dielectric (insulator) between the plates with some relative
permittivity, e_{r} = C/ C_{o},
where C is the dielectric capacitance value and C_{o} the vacuum value of capacitance. Then: C = e_{r}A/d and C_{o}
= e_{o }A/d, so that:

C/ C_{o}
= e A/d/ (e_{o }A/d) so: e_{r}** **= e_{ }/ e_{o }and
e = e_{r }e_{o.}

Therefore: C = A e_{r }e_{o} /d

For the capacitance of a parallel plate capacitor with dielectric between plates.

**b.
Capacitance of a Sphere.**

**Problem:
**Find the capacitance (general) for a given sphere, and the capacitance of the
Earth.

The electric potential in spherical geometry is:

f
= Q/ 4p e_{o}
a

Where r = a denotes the radius. Since the capacitance C = Q/V, then:

C
= Q/ [Q/ 4p e_{o} a ] =
4p e_{o} a

This shows the result for a sphere is not dependent on Q!

**For
the Earth:**

Since
the radius = 6 x 10^{6} m

**C
= **4p e_{o} r =
(1/9 x 10^{—9} F/m) (6 x 10^{6} m)
= 7 x 10^{-4} F

Or: 700 mF

c. **Energy
Stored in a Capacitor**.

Method (1): We
need V_{max} and V_{av}, which refer to the maximum potential
and the average.

V_{max} =
Q/C but V_{av} = V_{max} /2 =
Q/ 2C

The average work
per unit charge is: W _{av}/Q = V_{av }=

_{ }Q/
2C

**The
total work done: W = **V_{av}
(Q) – (Q/2C) Q = Q^{2}/2C

But Q = CV so: W =
(CV)^{2}/ 2C = ½ CV^{2}

Method (2): By Integration.

Energy stored = W
= ò_{0 }^{Vo} V(d Q) = ò_{0 }^{Vo} CV(dV)

But dQ = C dV so:
W = C ò_{0 }^{Vo} V dV = C[V^{2}/2]_{o}^{
Vo}

^{ }

W = CV_{o}^{2}/2
= ½ Q_{o}V_{o}

Note
here that Q_{o} is the final charge and V_{o} the final p.d**.**

**d.
Discharging a Capacitor.**

Consider the circuit shown in Fig. 6 below, wherein a capacitor C is discharged through resistor R. If the p.d. across C is V, then we have: Q = CV. The current discharged is then expressed:

I = - dQ/ dt

(i.e. Q, the charge, is decreasing as the current I increases)

**Fig. 6:
Basic Capacitative Circuit**

We know that Q = CV but by Ohm’s law: V = IR

So: Q = IRC = -CR (dQ/dt)

On separating variables we can integrate:

ò_{0 }^{t} dt/ CR = - ò_{Qo }^{Q} dQ/
Q

Where:

- ò_{Qo }^{Q} dQ/ Q = ò_{Q }^{Qo} dQ/ Q = ln (Q_{o}/Q)

**So
that: **

t/ CR =
ln (Q_{o}/Q)

**or:**

exp
(t/CR)** = **(Q_{o}/Q)

** Or:**

e
^{-(t/CR)} = Q/ Q_{o}

**Finally:**

**Q
= ***Q _{o} *

*e*^{-(t/CR)}

**e.
Capacitor Networks**

Capacitor networks
occur when multiple capacitors are connected in parallel or in series. A
parallel arrangement is shown in Fig. 7 below:

**Fig. 7:
Capacitors connected in parallel**

In
this case, the applied p.d. across each capacitor is the same but the charges
differ, according to the values C1, C2 and C3.

** **Thus: Q1 = VC1, Q2
= VC2 and Q3 = V C3

Then:

Q1 + Q2 + Q3 = V(C1 + C2 + C3) = Q

The single equivalent capacitor for this is then:

C = Q/ V = C1 + C2
+ C3

An arrangement of capacitors in series is shown in Fig. 8

**Fig. 8.
Capacitors connected in series.**

In this case, the potential difference divides up so all capacitors have the same charge but differing p.d.’s. Then:

V = V1 + V2 + V3 = Q/C1 + Q/ C2 + Q/ C3

And the equivalent series capacitor can be found from:

1/ C = V/ Q = 1/ C1 + 1/ C2 + 1/ C3

**Aside:**

** **In solar physics, a number of models for subflares involve weak capacitors - called *'weak* *double layers*' - in series, e.g**.**

**The basic model, depicted above, entails successive weak double layer (WDL) formation in a Vlasov-Maxwell (upper loop) plasma subject to anisotropic distribution of electrons in velocity space. The key point in the above model is that WDLs are needed to confine potential drops to extremely localized regions (in this case compact loops, also characterized by anomalous resistivity) with drops on the order of T**

_{e}/e where T

_{e}= kT

_{e}, in energy units. In at least one model, the potential drop associated with a double layer is obtained from:

V(t) = I dL(t)/dt
» 8.7
x 10 ^{9} V

For which an inductance drop is associated with the double layer. In this case the
*power* associated with the double layer would be:

P = I V(t)

A mono-energetic beam arises, with those
electrons of small pitch angle precipitating at loop foot points via the
loss-cone effect. The beam passes through turbulent wave fronts and weak double
layers - as refractive index changes-* en route* to the high beta regime. There
hard x-ray emission betrays the impact of the beam.

By way of greater generality, equivalent circuits using capacitors (and other components) can be used to represent simple loop flares, e.g.

Here, the simple loop contains L, C and R components. Can you write the series equation (in terms of the emf E)? Another simple loop equivalent circuit with a capacitor in parallel with an inductance is shown below:

**Suggested Problems:**

1) A capacitor of C = 500 mF is initially charged to a potential difference of 10.0 V and connected to a resistor R = 100 kW. Find:

a)
The
initial charge on the capacitor

**b)
**The
time constant for discharge through the resistor R.

**c)
**The
time for the initial charge to decay to one half its initial value.** **

d) The
energy stored in the capacitor by this time

2) Consider
two capacitors, C1 = 6mF and C2 = 3mF in series. If the operating
voltage is 18 V, find: i) the equivalent capacitance of the combination, ii)
the charge on each plate, and iii) the p.d.s across each capacitor**.**

3) A parallel plate capacitor has a capacitance of 0.001 mF.

a)
What
p.d. is required for a charge of 0.5 mC on
each plate?

**b)
**What
is the total stored energy?

**c)
**If
the plates are 1.0 mm apart, find the area of each plate.

4) In a particular solar loop there are** **10^{6} weak double layers in series. Each one is 10 cm across. If the total potential drop associated with *all the double layers* is DV » 10 ^{9} V, estimate: i) the potential difference across each, ii) the capacitance of each. (Assume the same charge Q = 2 Coulomb applies to each DL)

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