9C A
PHYSICS MIDTERM 1 May 15, 2006
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Problem 1 (2 points). If the potential profile has a form of stairs, at which points the electric field due to such profile reaches its largest values? A. At the bottom of stairs. B. At the top of stairs. C. At all points for which the potential jumps. D. At all points for which the potential remains flat. E. Electric field due to such profile will be constant and therefore does not reach its maximum anywhere. Solution: Since E(x)=-dV/dx, the electric field will go to infinity at each point where the potential jumps. Answer C.
Problem 2 (2 points). If you stretch a copper wire by 5 times what happens to its resistance? A Increases 5 times B Increases 25 times C. Decreases 5 times D. Decreases 25 times E. Does not change Solution: Since R=*L/A, the length L increases 5 times, and at the same time A decreases 5 times (to keep the volume of the wire V=LA constant), the resistance will increase 25 times. Answer B. Problem 3 (2 points). A charge is brought onto parallel plate capacitor, after which the distance between the plates is doubled. The voltage across the capacitor is A. B. C. D. E. Doubled Halved Does not change Increases 4 times Decreases 4 times.
Solution: The voltage across the capacitor V=Q/C. If the distance between the plates is doubled, Cnew=0*A/dnew = 0*A/(2d)=0.5 C, i.e. the capacitance is halved. As a result the voltage is doubled. Answer A.
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Problem 4 (2 points). A real battery is working in a circuit. Is it possible that the voltage across it becomes larger than its EMF? A. B. C. D. E. Impossible because the voltage of the real battery is always equal to its EMF Impossible because the voltage of the real battery is always smaller than its EMF Possible when the current through the battery is in the direction opposite to its EMF Possible when the current through the battery is in the same direction as its EMF Impossible because the real battery has so large resistance that the voltage is always zero.
Solution. The voltage across the real battery V=E-i*r where E is its EMF, i is the current inside it and r is its internal resistance. The formula is derived under the assumption that the current flows along the battery's EMF, i.e. i>0. If the current inside the battery flows in the direction opposite to its EMF, (for example because a larger battery of different polarity is connected in parallel with this battery) the current flows in the direction opposite to its EMF, and the voltage becomes V=E+i*r, i.e. larger than E. Answer C. Problem 5 (3 points). A positively charged particle (q=10 C, m=10 mg) is accelerated from the rest by an electric field existing in the space between points xi=0 and xf=10 cm. The electric field grows linearly as follows E(x)=E0*x, where E0=10 N/(mC*m). Find velocity of the particle when it exits such accelerator. Solution. First, let's first find the potential difference between points xf and xi: xf xf 1 1 x V ( x f ) - V ( xi ) = - E ( x )dx = - E0 xdx = - E0 x 2 |xif = - E0 ( x 2 - xi2 ) = f xi xi 2 2 1 - *10 *10+3 * (10 *10-2 ) 2 = -0.5*10+2Volts = -50Volts 2 Now the gain in the kinetic energy of each particle is a lost of its potential energy
mv 2 f / 2 - mvi 2 / 2 = - q[V ( x f ) - V ( xi )] = - qV and v f = ( -2 qV / m )1/ 2 = (2 *10 *10-6 *50 /(10 *10-6 ))1/ 2 = 10m / s
Problem 6 (3 points). Estimate the value of the force acting on a dielectric slab with =2 when it is inserted half-way into a parallel plate capacitor charged with Q=3 mC. Assume square plates of size 1 cm separated by distance 1 mm.
Hint. Evaluate the force as a derivative of potential energy of the capacitor .
Solution. When the dielectric is inserted into the capacitor, its capacitance changes. If each of the capacitor plate has length L, width w, and the dielectric is inserted onto distance x, we can look at this capacitor as two capacitors connected in parallel. The capacitance of the capacitor with dielectric is C1=*0*A1/d=*0*w*x/d while the capacitance of the remaining part is C2=0*A2/d=0*w*(L-x)/d. The total capacitance becomes x dependent and it is given by Ctot(x)=C1(x)+C2(x)=*0*w*x/d +0*w*(L-x)/d.
The potential energy of the capacitor which has a fixed charge Q on its plates is U(x)=Q2/[2Ctot(x)]. When dielectric is inserted the capacitance grows, therefore potential energy decreases. The force acting on a dielectric slab is the derivative of the potential energy
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dU Q2 d 1 Q2 d F ( x) = - Ctot ( x ) = =- = 2 dx 2 dx Ctot ( x ) 2C tot ( x ) dx Q2 2C 2 tot ( x ) [
0 w 0 w
d - d
]=
Q2
0w
2C 2 tot ( x ) d
[ - 1] =
Q 2Cair [ - 1] 2 LC 2 tot ( x )
When the capacitor is inserted half-way, x=L/2 and we have Ctot(L/2)=(+1)Cair/2. The force is given by L Q 2Cair 2Q 2 [ - 1] F( ) = [ - 1] = 2 2 L( + 1) 2 C 2 air / 4 LCair ( + 1) 2 Evaluating Cair=0*w*L/d=8.85*10-12*10-2*10-2/10-3=8.85*10-13 F Evaluating the force F=2*(3*10-3)2/(10-2*8.85*10-13)*(2-1)/(2+1)2=0.226 * 109 N
Problem 7 (3 points). A circuit is made of one 10 V battery and three capacitors with C1= 1pF, C2=3pF and C3=2pF. What is the charge accumulated on the plates of the capacitor 1?
Solution.
Since C1 and C2 are in parallel, their capacitance is C12=C1+C2= 4 pF. Let's evaluate total capacitance C123=1/(C12-1+C3-1)=1/(1/4+1/2)=4/3 pF. Total charge accumulated in C123 is Q123=C123*V=40/3 pC. Now note since capacitors C12 and C3 are in series, the same charge Q123 exists in C12 and in C3, i.e. Q123=Q12=Q3 Since we know the charge accumulated on C12 we can find the voltage across C12 as V12=Q12/C12=Q123/C12=40/(3*4)=10/3 V. Now we know the voltage across both C1 and C2 capacitors, so we can evaluate charges accumulated on each of them. In particular, the charge on C1 is Q1=C1*V1=C1*V12=1*10/3 =3.33 pC.
Problem 8. (3 points). A voltage V(t)=V0 exp(at) (V0=3 mV and a=3 s-1) is applied across a cylindrical wire having resistivity =10 *cm, diameter d=2 mm and length L= 31.4 cm. What is the value of charge passed thru this wire between time moments t=0 and t=1 seconds? Hint: Assume that the current at each given time moment t is i(t)=V(t)/R Solution: First estimate the resistivity R=*L/A=*L/(r2)=10*10-6*10-2*0.314/(3.14*10-6)=0.01
Second, estimate charge passing thru the wire as the integral of the current t2 1 t2 V t2 V Q = i (t )dt = V (t )dt = 0 e at dt = 0 [e at2 - e at1 ] t1 R t1 R t1 aR -3 3*10 ( e3 - e0 ) = 0.1* (2.713 - 1) = 1.89C = -2 3*10 3