1 of 8 CAPCT-1. A parallel-plate capacitor has square plates of edge length L, separated by a distance d. A second capacitor is made with L doubled and d decreased by a factor of 2. L d L L2L,dd/→→2 By what factor is the capacitance of the new capacitor increased? A) 1 B) 2 C) 4 D) 16 E) None of these. Answer: None of these. The capacitance increased by a factor of 8. The area (L×L) increased by a factor of 4. Another factor of 2 comes from the separation d. CapCT-2. A parallel-plate capacitor with capacitance C has a charge Q (meaning +Q on one plate, –Q on the other). The charge is doubled to 2Q (meaning +2Q on one plate, –2Q on the other). The capacitance C of the capacitor... A) doubled B) decreased by 2X C) remained constant D) None of these Answer: The capacitance remained constant. The ratio Q/V is fixed by the geometry (size, shape, separation) of the capacitor. If Q is increased, V also must increase, so the ratio Q/V stays the same. 9/29/2009 PHYS1120 Dubson Fa09 ©University of Colorado at Boulder2 of 8 CAPCT-3. A parallel-plate capacitor is charged up (+Q on one plate, –Q on the other). The plates are isolated so the charge Q cannot change. The plates are then pulled apart so that the plate separation d increases. The total electrostatic energy stored in the capacitor A) increases B) decreases + + + + + + + + + + + + + + + + + + + + +- - - - - - - - - - - - - - - - - - - - C) remains constant. (Hint: Did the person pulling the plates apart do positive work, negative work or no work?) As the plates were pulled apart, the energy density (energy per volume) = u = (1/2)εoE2 .... A) increased B) decreased C) stayed the same. Answer: The energy increased, which we can see in three different ways: Method I. The Q is fixed and the C decreases (since C = ε0A/d), so U = (1/2)Q2/C increases. Method II. The total energy stored in the field between the plates is (energy density) x volume = 122εoE Volume×bg. The field E = σ/εo does not change (since the charge and charge density σ do not change) but the volume between the plates increased, so the energy in the field increased. Method III. The external agent pulling the plates apart did positive work (since the plates attract). That positive work done was stored as increased electrostatic potential energy in the capacitor. The stored energy increased. 9/29/2009 PHYS1120 Dubson Fa09 ©University of Colorado at Boulder3 of 8 CAPCT-4. A parallel plate capacitor is attached to a battery which maintains a constant voltage difference V between the capacitor plates. While the battery is attached, the plates are pulled apart so their separation increases. The electrostatic energy stored in the capacitor A) increases B) decreases C) stays constant. Answer: Using U = (1/2)CV2, we can see that the stored energy decreased. The Q on the plates was not constant (since the plates are not isolated) so we cannot use a formula involving the charge Q. The voltage V between the plates is held constant by the battery. (That's what a battery does: it maintains constant voltage difference between its terminals.) The capacitance C decreased as the plates were pulled apart (by C = ε0A/d), so U = (1/2)CV2 decreased. Where did the energy go? Into the battery. What happened to the charge Q on the plates as the plates were pulled apart (at constant Voltage)? (Hint: What happened to the E-field? E-field is related to charge density.) Answer: The charge Q decreased. C = Q/V, since V was constant while C decreased, it must be that Q decreased. Alternatively, we can say that V = E d = constant, so if d increases, E must decrease (since V is constant.). If E decreases, the charge density must decrease (since E = σ/εo). If charge density decreases, the charge must decrease. CAPCT-5. A positive charge +Q and a negative charge –Q are held a distance R apart and are then released. The two particles accelerate toward each other as a result of their coulomb attraction. As the particles approach each other, the energy contained in the electric field surrounding the two charges.. A) increases B) decreases C) stays the same. Answer: The energy in the E-field decreases. The energy density of the E-field is 21o2Eε; this is NOT the energy, it is the energy per volume. If the E-field is constant, the total energy is 21o2Evolumeε×. If the E-field is not constant, the total energy is 21o2EdVε∫, where dV is a volume element. In this problem, the volume in which the E-field is large is roughly the space between the particles. This volume is decreasing rapidly as the particles approach each other. Although the E-field between the particles is increasing as they approach, the volume is decreasing more rapidly than the E-field is increasing, so the total energy is decreasing. Actually, + + + + + + + + + + + + + + + + + + + + +- - - - - - - - - - - - - - - - - - - - V+ 9/29/2009 PHYS1120 Dubson Fa09 ©University of Colorado at Boulder4 of 8 I am glossing over some details: the exact calculation is pretty tricky, due to the divergence of the E-field near the point charges. Another way to see this is : When the two particles are infinitesimally close, so that they "on top of each other", the (+) charge cancels the (–) charge, and the E-field is zero. Then there is no energy in the field, so the field energy must have been decreasing as they approached. Where did the field energy go? It goes into the increased KE of the particles as they accelerated toward each other, and it goes into light that is emitted. CAPCT-6. The inner surfaces of the two plates of a capacitor have uniform charge per area σ1 and σ2. Consider the gaussian surface S. σ1 S σ2 From Gauss's Law, you can conclude that A) σ1 = σ2 always. B) σ1 = – σ2 always. C) You cannot conclude anything from Gauss's Law) σ1 and σ2 can be anything Answer: σ1 = – σ2 Gauss's Law is enclosed0SQEdAε⋅ =∫KKv. We can argue that SEdA 0⋅ =∫KKv because (1) the top and bottom of the surface S is inside the metal where E = 0, and (2) on the sides of S, the E-field is perpendicular to the area vector EdA EdA0⊥⇒⋅=KKKK. From Gauss's Law, we conclude that Qenc = 0. Therefore the charge on the top plate must be the opposite of the charge on the bottom plate. Since the areas are equal, and the charges are
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