= (series) Capacitors in series. C eq. Hence. Capacitors in parallel. Since C 1 C 2 V 1 -Q +Q -Q. Vab V 2. C 1 and C 2 are in series

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1 Capacitors in series V ab V + V Q( + C Vab + Q C C C Hence C C eq eq + C C C (series) ) V ab +Q -Q +Q -Q C and C are in series C V V C +Q -Q C eq C eq is the single capacitance equivalent to C and C in series with the same charge Q Capacitors in parallel Since Q Q + Q C C ) ( + Q C+ C V C eq V ab V ab C Q C Q C eq +Q PHYS 53 8W -Q Here, QQ +Q

2 Hence C eq C + C (parallel) In fact, for n capacitors in parallel, C eq C C + C Energy storage in capacitors As a capacitor is being charged, it takes work to add successively more charge. What is that work? Where is the work done, or energy, stored? The first element of charge dq requires no work to be moved to the uncharged plate. Then as charge builds up, more and more work is required to add dq. At any time when the voltage is V, the work in moving dq is dq V. If the final charge is Q, and the final voltage V, then a reasonable guess for the work done is ½ QV. At an intermediate stage, the work done is W W dw C Q Q qdq C CV QV dw dqv qdq C These expressions give the work done in charging a capacitor, or the potential energy stored in a capacitor. PHYS 53 8W

3 Problem 3. Four electrons are located at the corners of a square,. nm on a side with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square? PHYS 53 8W 3

4 Q Note the analogy. The stored energy for a capacitor U cap is similar to C that for stored energy in a stretched spring Uspring kx Consider a parallel plate capacitor. The work done results in the creation of an electric field. Energy is stored in this field. CV The energy density u where d is the plate separation and A the area. Ad ε For a parallel plate capacitor C A and V Ed d Hence Dielectrics u ε E (energy density of any E field configuration in a vacuum). These are non-conducting materials used between conducting plates of a capacitor. Uses:. mechanically separates the conductors. increases max. V 3. increases C. Fig. 4.3 PHYS 53 8W 4

5 How does all this happen? Dielectrics alter the electric field in a capacitor (imagine parallel plates) because of polarisation. Some molecules (polar molecules) have asymmetrical charge distributions and are natural dipoles e.g. H O. With E, the molecules are oriented randomly as in Fig. 4.8 (a). When an electric field is applied, they show a preferential alignment as in Fig. 4.8 (b). Fig. 4.8 Even non-polar molecules will become weak dipoles in the presence of an electric field. These are induced dipoles (Fig. 4.9). The effect of the partial alignment of these dipoles is to produce a charge density at the surface of the dielectric. These charges are bound and not free to move around. Fig. 4.9 PHYS 53 8W 5

6 This redistribution of charges is called polarisation, and the material is said to be polarised. This polarisation of the dielectric sets up an opposing E field between the plates and this weakens the original E field. The original field is only partially cancelled (Fig. 4.). V Ed Since if E is reduced, so is V for a constant Q (Fig. 4.4). Fig. 4. If E, V, and C are the original electric field, voltage across the plates, and capacitance, then with the dielectric inserted, E K E (Q constant) K is the dielectric constant. V K V Hence (Q constant) Fig. 4.4 PHYS 53 8W 6

7 If Q is constant, and V changes, then the capacitance will change. For a decrease in voltage, there will be an increase in capacitance. C KC (definition of K) Representative values of K are shown. ( K ) Before a dielectric is inserted, E σ ε Afterward, E σ i ε σ ( σ i is the induced surface charge density on the dielectric). Taking a ratio of these two equations, we get for the induced surface charge density σ i σ( K ) PHYS 53 8W 7

8 From above, And hence CKC A A C Kε ε d d ε Kε is the permittivity of the dielectric (this is the definition). With the dielectric in place, the electric field energy density becomes u Kε E ε E Now let s look at Gauss s law for a dielectric. Since E lines are perpendicular, a suitable Gaussian surface is the box shown in Fig Only the side in the dielectric has an E flux through it. EA ( σ σi) A ε Fig. 4.3 PHYS 53 8W 8

9 Substituting for EA σa K ε σ i K Ed. A or Q ε KEA σa ε where Q is the enclosed free charge. Gauss s law in a dielectric. In this expression, note that has replaced KE E PHYS 53 8W 9

10 Problem 3.7 A solid sphere of radius R contains a total charge Q distributed uniformly throughout its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the self-energy of the charge distribution. (Hint: After you have assembled a charge q in a sphere of radius r, how much energy will it take to add a spherical shell of thickness dr having charge dq? Then integrate to get the total energy). PHYS 53 8W

11 Problem 4.5 In the figure, each capacitor has C4. µf And V ab +8.V. Calculate (a) the charge on each capacitor, (b) the potential difference across each capacitor, (c) the potential difference between points a and d.

12 Problem 4.6 The capacitors shown in the figure are initially uncharged and are connected with switch S open. The applied potential difference is V ab + V. (a) What is the potential difference V cd? (b) What is the potential difference across each capacitor after switch S is closed? (c) How much charge flowed through the switch when it was closed?