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54S08Exam1Solution

54S08Exam1Solution - Physics $4 80 Exam l x Q A ft 2 Q m...

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Unformatted text preview: Physics $4 80% Exam l x) Q A ft 2?; Q m “ti Section: Part l: Short Answer (3pm each) Circle the best answer: Q 1) Q 2 __--___.______-_____.®.--_ A positive charge q is located on a line midway between +2 C and — 3 C charges and is equidistant from each of them as shown. The net force on q from the two Charges a) has a negative x—component and a negative y-component. b) has a negative x—component and a positive y-component. has a positive X—component and positive y~component it has a positive x—component and a negative y—component. 2) A hollow conducting sphere with a net charge of Q has a charge —q inside located off center. The charge on the outer surface of the sphere is: Two graphite resistors with identical resistivity are shown. Which has the larger resistance? (numbers denote sizes in mm) _. 7 [email protected] Theleftone. A 2?; [2% = {y f b) The right one. c) Cannot determine. % 1 if g “3:. d) Both have the same resistance. 57‘ 4) Two large area insulators are uniformly charged with total charges as shown. The electric potential is highest at point b) b. c) c. d) cannot be determined for insulators. ‘ {,3 :2; :2’ #9 m i ”ataxia a? g»: 26: ‘ § 3:; egg 5) C , a The equivalent capacitance of the combination on the left is: a) 2C/3. b) 3C/4. @ 5C/3. ) C/2. 6) W I 5’59”???" (5%”? i 6W” T}??? ‘ISV 20v ”i5g&%et} at fig: Zaffitefl an??? The plate on the left is held at an electric potential of »15 V, while the conducting sphere on the right is held at + 20 V. A particle with a charge of q = —4 MC moves from the surface of the plate to the sphere. In doing so: «as— f ' i a) Itloses 140m ofkineticenergy. { 3 N VTf) flywieif b) There is no kinetic energy change. i It gains 140 tLl of kinetic energy. There is no change in potential energy since the particle moves horizontally. 7) fij‘mjgfwef An air—spaced capacitor with plates of area A and separation ti is charged as shown and disconnected from the battery. I put a dielectric plastic plate of area A, thickness d and dielectric constant K = 5 in between the plates. I let go of the dielectric plate, which is initially at rest. 3 It stays in because the stored energy is lower when it is in. ' It pops out to decrease the stored energy. c) The field in the region between the plates increases when the dielectric is added. d) The stored energy increases when the dielectric is added. ima—[WE-Wfl 8) Three identical light bulbs are connected as shown. When a current I = 1 A flows, 1 Watt of power is dissipated in bulb 3. The power dissipated in bulb 2 is a)2W. Parts afar/aw b)1W. *3 W. $2; Q g? 9 J; ay< " P2 ~53??? m .._. if f7” 9) One corner of a charged copper square is connected to a battery that maintains an electric potential of 15 V. The electric potential anywhere inside the plate: 0 is 15 V because the electric field is zero inside the plate. ) is 0 inside the plate. c) cannot be determined from the information given. d) depends on the distance from the surface of the plate. 10) Two open surfaces 1 and 2 bound a closed volume. The flux into surface 1 is —3 x 105 N—mZ/C. The flux out of surface 2 is +1 1 x 105 N— fill/C . The charge inside the volume is Q @i g? mg: 3 5.4%! an? a) 0. fig p negative. Positive. 6 2 if”? m a c the sign cannot be determined. 5 7? M g/& Part ll: Lon . Answer ———e ~1 11C +14 DC 1) (20 pts) Two charges sit on the x—axis separated by 2m as shown. A uniform field of SN/C is applied as shown. Use k = 9 x 109 ; l nC = 10'9 C. a) In terms of x and the given quantities, what is the magnitude and direction of the electric field from the —l nC charge at the point P? b) In terms of X and the given quantities, what is the magnitude and direction of the electric field from the +14 nC charge at the point P? c) Write the equation which determines the value of x for which the total electric field at the point P is zero, including the applied field. Guess the solution by assuming integer values of x. Xa 7% er» W ’ gym? We ”2 a? Ea 1-: r 6 We ,= r26 , ya}, i ”:13“? ”fl f? Etarfitx Z z‘t’q gar-t2}? K<1 2) (25 pts) An insulator sheet has a uniform charge density throughout its volume. The total charge is Q = + 3 x 10‘12 C and the thickness is 2d 2 0.02 m. The insulator is sandwiched symmetrically between two identical conducting plates 1 and 2, each with net charge zero. Both the insulator and each conductor have the same large surface area A = 3 m2 as shown. The dashed line denotes the x = 0 plane located in the center of the insulator. Use so = 8.85 x 10‘”. a) Explain in simple physical terms why the electric field E is zero at X = O. b) Using E = O in the plane at x = 0, find the total charge on the inner (facing the insulator) surface of conductor 2. c) Find the total charge on the outer (facing away from the insulator) surface of conductor 2. ae d) Find the magnitude and direction of the field t in the air space between the insulator and the conductor. %e X > Q u : 5.30 eeafif yflfflq- & ;@ ”5‘7“" gag fé‘fiwflp’ffi fig” f? :09 ~‘ Mane fievfiee =fl Q: ;@ (‘fiéficé&w/7fi@r .941 flan FE»? jaflgfi‘ff “755 Ea \ M if)? \— : .J g s; fim ‘1‘- lav-£13 if 3) (25pts) A circuit contains two batteries and five identical resistors with R = 2 Q. A current I flows through the lower left resistor. 21) Find the equivalent resistance Req of the four—resistor combination on the upper right. 13) Find the current I. c) What is the combined power dissipated in all of the resistors? I? i , e} fie? is ,fi 3s i gig? Q é} Meg? W2 wigs «=5 rigs”; “£9 7mg/Zffl EL? 35%;? fiflzr “Fez-fig“? 33% s? f:izaflriggfi ., ? 3 . ’ §d£‘=2326w zine? ...
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