In the post dated 5th August 2008, two free response practice questions on capacitors were given to you. As promised, I give below model answers along with the questions:
Two metal plates A and B, each of area 0.2 m2 are arranged parallel to each other as shown, with a separation of 1 cm in air. A regulated direct voltage source of output V volt remains connected between the plates so that there is an electric field of 2000 N/C in the region between the plates.
(a) Determine the output voltage V of the regulated voltage source.
(b) Determine the capacitance of the parallel plate capacitor formed by the plates A and B
(c) If the separation between the plates is gradually reduced (with the voltage source remaining connected between the plates), what quantities among the following will change? Put tick (√) marks against the changing quantities:
(i) Capacitance____ (ii) Charge on the plates____ (iii) Potential difference between the plates___
Give reasons for change (if any) and for no change (if any).
(d) During the decrease in the separation between the plates, will there be a current through the wires connecting the plates to the voltage source? Justify your answer.
(e) Determine the energy of the capacitor when the separation between the plates is reduced to half the initial value.
(a) The potential difference between the plates is equal to the otput voltage V of the voltage source. Since the electric field (E) between the plates is given by E = V/d where d is the separation between the plates, we have
2000 = V/10–2 from which V = 20 volt.
(b) Capacitance (C) of the parallel plate capacitor is given by
C = ε0A/d where A is the area of a plate.
Therefore, C = 8.85×10–12×0.2/10–2 = 1.77×10–10 farad.
[The value of ε0 will be given in your question paper].
(c) (i) Capacitance__√__ (ii) Charge on the plates__√__ (iii) Voltage between the plates____
The capacitance is given by C = ε0A/d where A is the area of a plate and d is the seaparation between the plates. Since the separation is decreased, the capacitance is increased.
The charge on the plates is given by Q = CV. The capacitance C is increased but the voltage V between the plates is unchanged (since the plates are connected across the fixed voltage source). The charge Q on the plates is therefore increased.
(d) During the decrease in the separation between the plates, the capacitance goes on increasing and hence more charges flow from the voltage source to the capacitor, satisfying the equation, Q = CV. So there is a charging current in the wires connecting the plates to the voltage source.
(e) The energy (U) of a capacitor of capacitance C charged to a potential difference V between its plates is given by
U = (½) CV2
Initially the capacitance of the capacitor made by the two plates is 1.77×10–10 farad as shown in part (a) above. When the separation (d) between the plates is reduced to half the initial value, the capacitance is doubled in accordance with the expression for capacitance, C = ε0A/d. Therefore, the energy is given by
U =(½) (2×1.77×10–10)×202 = 7.08×10–8 joule.
Question No.2 (For AP Physics C):
A spherical capacitor is made of two conducting spherical shells of radii r1 and r2 (r2 > r1). The inner shell has a total charge +Q and the outer shell has a total charge –Q. Now, answer the following questions:
(a) The space between the shells is filled with air of dielectric constant very nearly equal to 1. Calculate the potential difference between the shells.
(b) Determine the electric field at a point in between the shells, at distance a from the common centre of the shells.
(c) What are the values of the electric field and potential at a point outside both shells, at distance b (b > r2) from the common centre of the shells?
Justify your answer.
(d) Determine the potential at the common centre of the shells.
(e) The air in between the shells is replaced by a material whose dielectric constant has value K1 from r1 to R and value K2 from R to r2 ((r2 > R > r1). Show that the capacitance of this spherical capacitor is given by
C = 4πε0/[(1/K2R – 1/K1R) + (1/K1r1 – 1/K2r2)]
(a) The potential at all points inside the larger shell is the same as the potential at its surface. Therefore, the potential difference (between the shells) produced by the charge –Q on the outer shell is zero and we need consider the contribution (to the potential difference) by the charge +Q on the inner shell alone.
The potential (V1) of the inner shell due to its charge +Q is given by
V1 = Q/4πε0r1
The potential (V2) of the outer shell due to the charge +Q on the inner shell is given by
V2 = Q/4πε0r2
Therefore, potential difference V1 – V2 = (Q/4πε0)[(1/ r1) – (1/ r2)]
(b) The electric field at points inside the outer shell due to the charge on it is zero (since there is no potential gradient because of its charge). The field (E) in between the shells at distance a is therefore due to the charge +Q on the inner shell alone and is given by
E = Q/4πε0a2
This field is directed radially outwards.
(c) The electric potential and field at all points outside both shells (at distance b > r2) is zero since the potential and field at points outside any charged spherical shell is produced as though the entire charge is concentrated at the centre. We have therefore positive and negative charges of equal magnitude and the net charge imagined to be concentrated at the centre is zero.
[The electric field can be proved to be zero by applying Gauss’s law, considering a Gaussian spherical surface of radius b passing through the point. The net charge enclosed by the gaussian surface is zero so that the field is zero].
(d) The potential (V) at the common centre of the shells is the sum of the potentials due to the charges on the shells and is given by
V = (1/4πε0)[(Q/r1) – (Q/r2)] = (Q/4πε0)[(1/ r1) – (1/ r2)]
(e) The capacitance (C) is given by
C = Q/V where V is the potential difference between the shells.
But V = V1 – V2 where V1 and V2 are respectively potentials on the inner and outer shells with the composite dielectric in between them.
If E is the electric field at a point in between the shells, at distance r from the centre, we have
V = – r2∫r1 Edr = r1∫r2 (Q/4πε0K r2)dr where K is the dielectric constant which varies with r (K has value K1 from r1 to R and value K2 from R to r2).
Therefore, V = (Q/4πε0) r1 ∫r2 dr/ Kr2
= (Q/4πε0) [{– 1/ K1r}R r1 + {– 1/ K2r}r2R]
= (Q/4πε0) [(1/K2R – 1/K1R) + (1/K1r1 – 1/K2r2)]
The capacitance is given by C = Q/V = 4πε0/[(1/K2R – 1/K1R) + (1/K1r1 – 1/K2r2)]
[You can obtain the above expression from the expression for the capacitance of a spherical capacitor with a material of dielectric constant K, given in the post dated 14th July 2008:
C = 4πε0K ra rb /(rb – ra)
The radii ra and rb are to be replaced by r1 and r2. Further, C will be the series combined value of two spherical capacitors C1 and C2 where C1 is made by spherical shells of radii r1 and R, with material of dielectric constant K1 in between and C2 is made by spherical shells of radii R and r2, with material of dielectric constant
Since you have to borrow the expression for the simple spherical capacitor, this method will not be as rigorous as the one we used above]
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