Basic Electrostatics - An Introduction (Part 2)

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_rA/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 10⁶ m

C = 4p e_o  r = (1/9 x 10^—9 F/m) (6 x 10⁶ 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 = Q²/2C

But Q = CV so: W = (CV)²/ 2C = ½ CV²

Method (2): By Integration.

Energy stored = W =  ò₀ ^Vo  V(d Q) = ò₀ ^Vo  CV(dV)

But dQ = C dV so: W = C ò₀ ^Vo  V dV = C[V²/2]_o ^Vo

 

W = CV_o²/2 =   ½ Q_oV_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:

  ò₀ ^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 10⁶ 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)