9C
MIDTERM PHYSICS 1
February 13, 2008
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Problem 1. (2 points) Three very large parallel uniformly charged non-conducting sheets with the surface density =+8.85x10-12 C/m2 are placed at the distance L=5 cm one from each other. What is the value of electric field between the sheets, outside the sheets? A) B) C) 0 N/C between all sheets, 1.5 N/C outside. 1.5 N/C between all sheets, 0 N/C outside. ? 1.5 N/C before first sheet, 1 N/C between first and second sheet, 0.5 N/C between second and third sheet, 0 N/C beyond third sheet. 1.5 N/C before first sheet, 0.5 N/C between first and second sheet, 0.5 N/C between second and third sheet, 1.5 N/C beyond third sheet. 0 N/C before first sheet, 0.5 N/C between first and second sheet, 1 N/C between second and third sheet, 1.5 N/C beyond third sheet. 1 2 3 ? ? ? + + +
D)
E)
Solution: Each sheet produces electric field directed to the left from its left, and directed to the right from its right. The value of the electric field from each sheet is /20=0.5 N/C. From the very left and very right, all three sheets contribute to the field, therefore E=3*0.5=1.5 N/C. In the region between sheet 1 and 2 as well as between 2 and 3 contributions from 2 sheets cancel, therefore E= 0.5 N/C. Answer D.
Problem 2. (2 points) A charge Q is placed on a solid metal cylinder as shown. How will this charge be distributed after a while? Q A) The charge will be distributed uniformly over the volume of the cylinder. B) The charge will be distributed over the surface of the cylinder. C) The charge will be concentrated as a point charge at the center of the cylinder. D) The charge will continuously move along the surface, creating a constant current. E) The charge will be distributed over the tubular part of the cylinder only, but not the ends.
Solution: Since we are talking about metal, i.e. conducting cylinder, the charge will be distributed over its surface only. Answer B
Problem 3. (2 points) An electron goes from one equipotential surface to another along one of the four paths shown below. Rank the paths according to the work done by the electric field, from least to greatest. A. B. C. D. E. 4, 3, 2, 1 4 and 2 tie, then 3, then 1 4, 3, 1, 2 1, 3, 4 and 2 tie 1, 2, 3, 4
Solution: The work done by electric field in cases 1,2,3,4 is W1=|e|*30 V, W2=|e|*10 V, W3=|e|*20 V, W4=|e|*10 V, where |e| is the absolute value of the charge of the electron. Therefore, W2=W4 is the least work, then W3, and W1. Answer B.
Problem 4. (2 points) A parallel plate air capacitor has a dielectric slab partially inserted inside it. The capacitor has charge Q on its plates. Is there a force experienced by the slab? +Q A. The capacitor is pulling the slab further inside. B. The capacitor is pushing the slab out. C. The slab is attracted to the left plate of the capacitor D. The slab is attracted to the right plate of the capacitor E. There is no force acting on the slab. C Air -Q
Solution: The potential energy of the capacitor is a function of its capacitance, while the force acting on the slab is a derivative of the potential energy. Since the capacitor has charge Q, the potential energy of the capacitor U=Q2/2C, i.e. reversely proportional to its capacitance. To reach the equilibrium, the system tries to minimize its potential energy, which can be achieved by pulling the slab inside: in this case, capacitance the of the capacitor increases and leads to the decrease of the potential energy according to U=Q2/2C. Answer A.
Problem 5. (3 points) Three charges are arranged in the corners of the square as shown in the figure. What is the absolute value of electric field made by those charges at the bottom right corner if Q=1 mC and each side of the square is 1 cm? Draw the direction of this electric field. Answer in units N/C. y Solution: Electric field is a superposition of the fields made by three charges. 1. Top left charge produces the field E1x=+|E1|cos(45)=+kQ/(1.41a)2 /1.41 = +kQ/a2/2/1.41 E1y=-|E1|sin(45)=-kQ/(1.41a)2 /1.41 = -kQ/a2/2/1.41 2. Top right charge produces the field E2x=0 E2y=+|E2|= +kQ/a2 3. Bottom left charge produces the field E3x=-|E3|= -2kQ/a2 E3y=0 The total field is a superposition of all individual fields Etot,x= E1x+ E2x+ E3x=+kQ/a2/2/1.41-2 kQ/a2 =-1.64 kQ/a2 Etot,y= E1y+ E2y+ E3y=-kQ/a2/2/1.41+kQ/a2 =+0.65 kQ/a2 The absolute value of the field is |Etot|=(Etot,x2+Etot,y2)1/2= kQ/a2*(1.642+0.652)1/2=8.988*109*10-3/10-4*1.764=15.85*1010 N/C. -2Q Etot E3 E2 x +Q -2Q |E| - ? x +Q -Q
y -Q
E1
Problem 6. (3 points) A charge q=1 C of mass m=1 mg is hung on a non-conducting thread and is placed inside a parallel plate capacitor with the distance between the plates d=10 cm. When capacitor is connected to the 10 V battery, the charge deviates from the equilibrium due to an electric field that appears between the plates as shown in the figure. Find the angle of deviation . (Assume acceleration of gravity g=10 m/s2)
Solution: First estimate the electric field in the capacitor E=V/d=10/0.1=102 N/C. Then there is a force acting on the charge from the electric field F=qE directed horizontally. On top of that, the charge feels the gravity mg pointing downwards as well as tension T pointing along the thread. The balance of forces reads qE+mg+T=0 Along vertical axis we have mg=Tcos() Along horizontal axis we have Tsin()=qE. Therefore, mg tan() = qE tan()=qE/mg= 10-6*102/(10-6*10)=10. =84 degrees ground V=10 V
-?
q=1 C
Problem 7. (3 points) The following circuit is made of the capacitors C1=2 pF, C2=4 pF, C3=6 pF, C4=8 pF, and the 10 V battery. a. What is the equivalent capacitance in this circuit? b. What is the voltage across the plates of the capacitor C3? c. What is the charge accumulated on the capacitor C2? C1 Solution: a. C2 and C3 are in parallel, therefore their resulting capacitance C23=C2+C3=10 pF. After that C1, C23, C4 are in series, the total equivalent capacitance is Ceq=(1/2+1/10+1/8)-1=1.379 pF b. The charge accumulated on the capacitors C1, C23, and C4 is the same and is equal to the charge accumulated on Ceq. This charge q=Ceq*V=13.79 pC. This charge is accumulated on the plates of two capacitors C2 and C3 together. Since C23=10 pF, the voltage across C23 can be found as V23=q/C23=13.79/10=1.379 V. This is the same voltage across the capacitor C3. c. The charge on the plates of the capacitor C2 is C2*V2=C2*V23=4*1.379=5.517 pC C3
C2 C4
V
Problem 8. (3 points) An insulating sphere with radius R shown in the figure below has a charge Q uniformly distributed over its volume. Using the Gauss Law derive the expression for the electric field E as a function of distance r from the center of the sphere when a) r<R and b) r>R. c) Plot the electric field E(r) as a function of r.
r
Solution: Considering the Gaussian surface as a sphere of radius r<R, the total flux =E(r)*4r2 is equal to the charge enclosed inside the sphere: q(r)/0=QV(r)/V(R)=Q(r/R)3 where V(r)=4/3 r3. a. We have for r<R E(r)=Q/(40r2)*(r/R)3 = Q/(40R3)*r which is just a linear function of r. b. For r>R we just enclose the full charge Q, therefore the electric field E(r)=Q/(40r2) is similar to the point charge placed at the origin. c. The plot E(r) is thus linear function of r until r=R and then falling off as 1/r2.