Explore BrainMass

Explore BrainMass

    Capacitors: Types, Expressions, Effect of Insertion

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    What is a capacitor? Describe different types of capacitors and derive expressions for their capacitances. Derive expression for the energy stored in a parallel plate capacitor. Discuss the effect of insertion of a dielectric medium between the plates of a capacitor.

    © BrainMass Inc. brainmass.com September 29, 2022, 9:32 pm ad1c9bdddf
    https://brainmass.com/physics/gauss-law/capacitors-types-expressions-effect-insertion-585986

    SOLUTION This solution is FREE courtesy of BrainMass!

    See the attached file.

    Understanding Capacitors

    1. What is a capacitor?:

    Capacitor is a device which stores electrostatic energy and is widely used in electric and electronic circuits. In general, a capacitor is an arrangement of two conducting bodies, called plates separated by a small distance between them. In principle these bodies can be of any shape (though in practice simple shapes such as plates, spheres or cylinders are used). The gap between the two bodies can have air or any other dielectric medium. If a charge +Q is given to one of the plates and -Q to the other (say by connecting the two plates to a battery momentarily), the charges will spread themselves on the surfaces of the respective plates. The plates will be equipotential surfaces, each having a voltage (say V1 and V2). There will thus be a potential difference V = V1 - V2 between the plates and an electrostatic field will be set up between them.

    2. Capacitance of a capacitor:

    When we put a certain charge +Q on one of the plates and -Q on the other, the potential difference that develops between the plates shall have a unique value for a given capacitor and the same is determined by the construction and dimensions of the capacitor. Thus we can say that the ratio of the charge on each plate and the voltage developed between the plates is a constant for a given capacitor. This constant has been given the name "Capacitance of the Capacitor" and it is represented by symbol C. Thus

    Capacitance C = Q/V Coulombs/Volt

    The units of capacitance are Coulombs per Volt and the same is also called Farad. Thus 1 Coulomb/Volt = 1 Farad

    For a given potential difference between the plates, higher the charge Q that can be stored on the capacitor plates, higher the capacitance C. Thus, capacitance is a measure of the charge storing capacity of a capacitor.

    3. Capacitance of a parallel plate capacitor:

    Let us consider a capacitor comprising of two parallel metallic plates each with area A and separated by a distance d. Let the charge density on each plate be σ c/m2 (on one plate +ve and on the other -ve) and potential difference between the plates V.

    +σ c/m2 Area A
    Pot. Diff. V
    Gap d
    -σ c/m2

    The electric field intensity between two infinite plates each of which carries a charge density of σ (one plate +ve and the other -ve) is given by :

    E = σ ...(1)
    ε0

    Though the above expression is for infinite sheets, we can use the same approximately if the dimensions of the sheets are >> d (i.e.A >> d2).

    Since electric field intensity between the plates is also defined as the potential gradient, E = V/d ...(2)

    And total charge Q on each plate = σ . A => σ = Q / A ...(3)

    Substituting above values for E and σ in (1) we get :

    V = Q => Q = ε0 A
    d Aε0 V d

    By definition Q/V is the capacitance; hence, capacitance of a parallel plate capacitor is given by: C = ε0 A/d ...(4)

    4. Capacitance of a spherical capacitor:

    Let us consider a spherical capacitor comprising of two concentric, hollow spheres with radius ra and rb. Let the outer sphere be given +Q charge and the inner sphere -Q charge which spread uniformly on the respective sphere surfaces.

    +Q

    rb
    -Q
    ra

    We know the potential of a sphere (on the surface and inside) of radius R carrying a charge Q is given by: V = Q/4Πε0R and outside at a distance r from the centre: V = Q/4Πε0 r

    Thus, potential of inner sphere for r < ra (surface and inside) Va = - Q/4Πε0ra and for r > ra , Va= - Q/4Πε0 r .

    Potential of outer sphere for r < rb (surface and inside) Vb = Q/4Πε0rb and for r > rb , Vb= Q/4Πε0 r .

    Thus, net potential on the surface and inside the inner sphere (r < ra) = Q/4Πε0rb - Q/4Πε0ra = Q/4Πε0 (1/rb - 1/ra).

    Net potential at a distance r from the centre, outside the outer sphere (r > rb) = Q/4Πε0r - Q/4Πε0r = 0

    Net potential in the space between the two sphere (ra < r < rb) = Q/4Πε0rb - Q/4Πε0r = Q/4Πε0 (1/rb - 1/r).

    Thus, in the space between the two spheres, the potential varies between Q/4Πε0 (1/rb - 1/ra) (for r = ra at the inner sphere) and zero (for r = rb at the outer sphere).

    The potential difference between the two spheres = V = Q/4Πε0 (1/rb - 1/ra)

    Capacitance of a spherical capacitor C = Q/V = 4ε0 (ra.rb/(rb - ra) ...(5)

    5. Capacitance of cylindrical capacitor:

    Let us first derive a general expression for field in a hollow charged cylinder with radius R and length l and charge density σ.

    R

    Charge density +σ
    Gaussian
    l surface r>R

    Let us first consider as Gaussian surface a cylinder with radius r such that r > R. As per Gauss theorem we can write :

    E (r) S = Q/ε0 => E (r) 2r.l = Q/ε0

    E (r) = Q/2ε0 l r (direction of field vector being radially outwards)

    Next we consider the case of r < R. Since the charge enclosed within the Gaussian surface is zero, the field with in the cylinder shall be zero.

    Let us consider a capacitor with two concentric hollow cylinders as plates with radius ra and rb. Let the inner cylinder be given a charge +Q and the outer one -Q.

    Outer cylinder radius rb.
    Charge -Q
    Inner cylinder radius ra
    Charge +Q

    Based upon the above we can conclude:

    For r < ra , the field because of both cylinders shall be zero.

    For ra < r < rb the field because of outer cylinder shall be zero where as same due to
    the inner plate shall be:

    E (r) = + Q/2 ε0 l r

    For a varying electric field, the electric field E, potential difference V are related as:

    E = dV/dr (electric field is the potential gradient)

    dV = E(r) dr = - (Q/2ε0 l r) dr

    Integrating we get:
    rb
    V = (Q/2ε0 l)  1 dr
    ra r

    V = (Q/2ε0 l)[ln rb - ln ra] = (Q/2ε0 l)[ln(rb/ra)]

    Capacitance Q/V = C = 2Πε0l/[log(rb/ra)] ...(6)

    6. Equivalent capacitance of i) capacitors in series, ii) capacitors in parallel:

    i) Capacitors in series: Let us consider the case of three capacitors in series as shown and see what is the equivalent capacitance of this combination.

    + V1 - + V2 - + V3 -
    +Q -Q +Q -Q +Q -Q

    C1 C2 C3

    + V -

    Let us assume the left plate of the extreme left capacitor and the right plate of the extreme right capacitor are connected to a battery to charge the capacitors. Let us assume that a charge +Q is acquired by the left plate of the left most capacitor from the positive terminal of the battery and -Q by the right plate of the right most capacitor from the negative terminal of the battery. It can be seen that the other plates of the capacitor combination will acquire charges as shown through induction. These plates acquire same charge Q with polarities as shown.

    Let the capacitances of the three capacitors be represented by C1, C2 & C3. Then, the potential drops across each of the capacitors shall be:

    V1 = Q/C1 , V2 = Q/C2 , V3 = Q/C3

    Adding we get for total potential drop across the three capacitors:

    V = V1 + V2 + V3 = Q [1/C1 + 1/C2 + 1/C3]

    We can view the above expression as that of a single capacitor with effective capacitance C such that 1/C = 1/C1 + 1/C2 + 1/C3 in which case we get: V = Q / C

    Hence, the effective (or equivalent) capacitance of n capacitors with individual capacitances C1, C2, ......., Cn in series is given by:

    1/C = 1/ C1 + 1/ C2 + ...... + 1/ Cn ...(7)

    ii) Capacitors in parallel : Now let us consider the case of three capacitors in parallel as shown.

    +Q1 -Q1

    C1
    +Q2 -Q2

    C2

    +Q3 -Q3
    C3

    + V -

    Since the left and right plates of all the capacitors are connected as shown, the potential drop across each capacitor has to be same say V. If the capacitances of the three capacitors be different represented by C1, C2, C3, each one of them will have an amount of charge determined by the potential V and the capacitance. Let these charges be represented by Q1, Q2, Q3.

    Hence, we can write: Q1 = C1.V Q2 = C2.V Q3 = C3.v

    For total charge stored we can write : Q = Q1+Q2+Q3 = V[ C1 + C2 + C3 ]

    Above expression can be viewed as representing an equivalent capacitor with capacitance [C1 + C2 + C3] with potential drop V across it and storing a total charge Q = Q1+Q2+Q3 .
    Hence, the equivalent capacitance of n capacitors in parallel is given by:

    C = C1 + C2 +.......+ Cn ...(8)

    7. Energy stored in a parallel plate capacitor:

    Let us first see how the energy (potential) builds up in the capacitor. Let us assume the charge on the +ve plate is built up by progressively transferring small amounts of charge Q from the negative plate to the positive plate. As the amount of charge increases the potential difference also increases. As such transfer of each successive installment from the negative plate to the positive will have to overcome higher repulsive force thereby doing positive work which is stored as potential energy in the capacitor.

    Let us assume a situation when charge q has already been transferred and we transfer the next installment of q. The potential difference V with charge q will be V = q/C .

    By definition potential difference V is the work done in moving a unit positive charge from the -ve plate to the +ve plate. Thus the work done w in moving a charge q from -ve to +ve will is given by:

    w = V q = (q/C) q

    As q approaches infinitesimally small values we can write the above equation as:

    dw = 1 q.dq
    C
    Q
    Integrating we get, w = 1  q.dq = 1 (Q2/2) = 1 CV2 (since Q = CV)
    C 0 C 2

    Thus the energy stored in a parallel plate capacitor is given by w = Q2 = CV2 ...(9)
    2C 2

    Total energy in a combination of capacitors (series or parallel) is given by the sum of the energies stored in individual capacitors.

    8. Energy density of capacitor:

    As we have seen above, energy stored in a parallel plate capacitor = CV2 ...(10)
    2
    Capacitance for parallel plate capacitor C = aε0A/d

    Further, since potential difference V can be written as V = E.d.

    Replacing C and V in (10) we get :

    Energy stored in a parallel plate capacitor = CV2 = 1 [ε0 A] [E.d]2
    2 2 d

    = 1ε0 E2 (A.d)
    2
    Since (A.d) in the volume between the plates of the capacitor, we can write above expression as:

    Energy Stored = Energy density = ε0 E2/2 ...(11)
    Volume

    This expression can be states as: wherever there is a non zero electric field E, there is an electrical energy density (i.e. energy per unit volume) of the magnitude ε0E2/2 Jm-3

    9. Common potential and total energy of capacitors connected in parallel:

    Let us consider a capacitor with capacitance C, charged with a charge Q it acquires a voltage V. We can write V = Q/C

    Let it now be connected in parallel to another identical uncharged capacitor. Some of the charge from the first capacitor will flow to the second one. Let us assume after the steady state has been achieved, the charges stored in the two capacitors are Q/ each (thus Q = 2Q/) and the common voltage across the combination of capacitors is VC.

    We can write for each capacitor:

    Q/ = VC C

    Q = 2Q/ = 2 VC C = VC

    From above we get VC = V/2 ...(12)

    ii) Total energy of the first capacitor before it is connected to the second one is
    given by w = 1 CV2 .
    2
    As we have seen, after connection of the two capacitors, the common voltage VC is equal to V/2. Thus the total energy of the two capacitors is equal to:

    wC = 2 x 1 C(V/2)2 = C V2 = w
    2 4 2
    Thus the energy of the combination is one half of that of the single capacitor. Where does the balance energy go? As soon as we connect the second uncharged capacitor in parallel to the first, a transient state sets in. In this transient state the charge in the first capacitor steadily decreases and that in the second rises till a steady state is reached. Since during the transient period the current in the circuit is varying, energy is radiated in the form of electromagnetic radiation. Also some of the energy is lost as heat.

    10. Effect of dielectric between the plates of a capacitor (instead of vacuum) on its capacitance.

    Let us consider a parallel plate capacitor with charge Q and potential V. For vacuum as medium between the plates the capacitance is given by :

    C0 = ε0A/d where A and d are the area of each plate and separation respectively and ε0 is the permittivity of vacuum.

    Now let us assume another dielectric medium is introduced in between the plates. There is an electric field between the plates pointing from the positively charged plate towards the negative. Under the influence of electric field the atoms & molecules of the dielectric get polarized because the positively charged nucleus gets slightly shifted in the direction of the field, whereas the negatively charged electrons get shifted opposite to the electric field. This way each atom/molecule behaves like a dipole.

    Now if we consider a small Gaussian surface any where inside the dielectric, the net field through the same shall be zero as the net charge enclosed is zero (+ve and -ve charges being equal and opposite). However, at the surface of the dielectric, the situation is different. At the surface towards the head of the field arrow, one layer of atoms/molecules dipoles will form a layer of +ve charge. Similarly at the opposite surface a layer of negative charge will be formed. These layers of positive and negative charges will produce a field which, as we can see will oppose the field E0 in vacuum. Due to polarization of the dielectric, the field between the plates is effectively reduced from its value for vacuum. That means the reduction in the potential difference between the plates for the same charge.

    Since capacitance is given by the expression C = Q/V, reduction in the value of V for same charge Q means increase in the capacitance value of the capacitor. As we know capacitance in vacuum for a parallel plate capacitor is given by: C0 = ε0 A/d ...(13)

    Let the permittivity of the dielectric medium be ε. Then the capacitance with the dielectric medium C is given by :

    C = ε A/d ...(14)

    From (13) and (14) we get: C = (C0/ε0) ε = K C0 ...(15)
    where K = ε/ε0 is called relative permittivity or the dielectric constant of the medium. K for vacuum is 1 and its value for any other medium is greater than 1. For example for air it is 1.006, for glass 3-4, for transformer oil 2.24 and for water it is 80.

    Equation (15) suggests an alternative definition of dielectric constant of a medium as the ratio of capacitances of a capacitor with the dielectric medium between the plates and with the vacuum between the plates.

    11. Effect of dielectric between the plates of a capacitor (instead of vacuum) with constant charge, on energy stored.

    Let a capacitor (with vacuum as medium) be connected to a battery. After full charging of the capacitor to charge +Q let the battery be removed. The charge given to the capacitor is now fixed. Now let a dielectric medium of dielectric constant K be introduced between the plates.

    Energy stored in the capacitor is given by Q2. Substituting C = KC0 we get:
    2C
    Energy stored with dielectric medium = Q2
    2KC0
    Thus for a given charge on the capacitor, the energy stored in the capacitor with the dielectric medium is less than the energy stored with vacuum by a factor of 1/K.

    12. Effect of dielectric between the plates of a capacitor (instead of vacuum) with constant voltage, on energy stored.

    Let a capacitor (with vacuum as medium) be kept connected to a battery, keeping the voltage applied to the capacitor constant. Now let a dielectric medium of dielectric constant K be introduced between the plates.

    Energy stored in the capacitor is given by CV2/2. Substituting C = KC0 we get:

    Energy stored with dielectric medium = KC0V2/2

    For a given charge the energy stored in the capacitor with the dielectric medium is greater than the energy stored with vacuum by a factor of K.

    13. Capacitance of a parallel plate capacitor with a conducting slab (of thickness less than the gap between the plates) introduced between the plates.

    Let us consider a parallel plate capacitor with area of the plates A and separation between the plates d. Let the charge on the plates be +Q and potential difference V. Now let a conducting slab of thickness t be introduced between the plates.

    +Q Area A
    Ev
    -Q
    d t + + + + + + + + + + E = 0
    +Q
    -Q

    The face of the slab closer to the positively charged capacitor plate will acquire -ve charge (-Q) by induction and the other face will acquire +ve charge (+Q). If the potential difference between the plates be V, then we can write :

    V = Vv + Vs where Vv is the potential drop across the vacuum portion of the gap between the plates (i.e. thickness d-t) and Vs is the potential drop across the slab (thickness t).

    As we know, a conductor is an equipotential mass. Thus, there is no potential drop across the conducting slab i.e. Vs = 0. Thus, we get :

    V = Vv ...(16)

    We know value of electric field intensity Ev in the vacuum portion of the gap as

    Ev = σ = Q ...(17)
    ε0 ε0A

    Since E = Potential gradient, we can write from (16) & (17):

    Ev = Vv /(d-t)

    => Q = V
    ε0A d-t

    => C = Q/V = ε0A/(d-t) ...(18)

    Since capacitance with only vacuum between the plates is given by C0 = ε0A/d, (18) can also be written as: C = C0d/(d-t) = C0/(1-t/d) ...(19)

    Obviously, C > C0.

    The effect of introducing a conducting slab between the capacitor plates is to increase its capacitance.

    Not: Another way to look at the above is as follows: We can treat the single capacitor with a conducting slab in the gap as equivalent to two capacitors, each with a gap of (d-t)/2 in series (for simplicity the conducting slab is assumed to be placed in the middle of the gap). Thus, each capacitor has a capacitance of 2ε0A/(d-t). The equivalent capacitance of the two such capacitors in series is ε0A/(d-t).

    14. Capacitance of a parallel plate capacitor with a dielectric slab introduced between the plates.

    Let us consider a parallel plate capacitor with area of the plates A and separation between the plates d. Let the charge on the plates be +Q and potential difference V. Now let a dielectric slab of thickness t be introduced between the plates.

    +Q Area A
    -q Ev

    d t + + + + + + + + + + Es
    +q
    -Q

    The face of the slab closer to the positively charged capacitor plate will acquire -ve charge and the other face will acquire +ve charge on account of polarization of the molecules of the dielectric (as explained earlier). If the potential difference between the plates be V, then we can write :

    V = Vv + Vs ...(20)

    where Vv is the potential drop across the vacuum portion of the gap between the plates (i.e. thickness d-t) and Vs is the potential drop across the slab (thickness t).
    We know value of electric field intensity Ev in the vacuum portion of the gap as Ev = σ = Q ...(21)
    ε0 ε0A

    and in the dielectric medium as Es = σ = σ = Q ...(22)
    ε Kε0 Kε0A
    Where ε and K are the permittivity and dielectric constant of the medium respectively.

    Since E = Potential gradient, we can write from (20), (21) & (22) :

    V = Ev (d-t) + Es t

    => V = Q (d-t)/ ε0A + Q t/Kε0A

    => V = [Q/ ε0A] [(d-t) + t/K] = [Q/ ε0A] [d - t(1- 1/K]

    => C = Q/V = ε0A/[d - t(1- 1/K] ...(23)

    Since capacitance with only vacuum between the plates is given by C0 = ε0A/d, (23) can also be written as :

    C = C0d/[d - t(1- 1/K] = C0/[1 - (t/d)(1- 1/K] ...(24)

    Obviously, C > C0.

    Thus, the effect of introducing a dielectric slab between the capacitor plates is to increase its capacitance.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com September 29, 2022, 9:32 pm ad1c9bdddf>
    https://brainmass.com/physics/gauss-law/capacitors-types-expressions-effect-insertion-585986

    Attachments

    ADVERTISEMENT