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 L d L 2L d d 2 A 1 B 2 By what factor is the capacitance of the new capacitor increased 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 Boulder 2 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 2 density energy per volume u 1 2 oE 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 0A 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 1 o E 2 Volume The field E o does not change since the charge and charge density 2 do not change but the volume between the plates increased so the energy in the field increased b g 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 Boulder 3 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 V 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 0A 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 o 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 12 o E 2 this is NOT the energy it is the energy per volume If the E field is constant the total energy is 2 2 1 1 2 o E volume If the E field is not constant the total energy is 2 o E dV 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 9 29 2009 PHYS1120 Dubson Fa09 University of Colorado at Boulder 4 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 1 and 2 Consider the gaussian surface S 1 S 2 From Gauss s Law you can conclude that A 1 2 always B 1 2 always C You cannot conclude anything from Gauss s Law 1 and 2 can be anything K K K Qenclosed We can argue that E dA 0 v 0 S S because 1 the top and bottom of the surface S is inside the metal where E 0 and 2 on the K K K K sides of S the E field is perpendicular to the area vector E dA E dA 0 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 equal sized the charge per areas must also be equal sized but opposite in sign Answer 1 2 Gauss s Law is K v E dA 9 29 2009 PHYS1120 Dubson Fa09 University of Colorado at Boulder 5 of 8 CAPCT 7 Consider a metal in electrostatic equilibrium The voltage difference between two different points in the metal A is zero always B could be positive negative …
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