General Physics II
Name PHY 112
General Physics II
Solutions Spring 2003 Dr. Satinder S. Sidhu
Test 3 Questions 1 through 3 all refer to the situation in the following figure. It shows three charges placed on the horizontal axis and two empty points A and B on the vertical axis. Unless directed otherwise, you should include the effects of all three charges in all calculations.
1. Calculate the electric field at the point A. Clearly state its magnitude and direction.
20 points
In the following, I refer to the leftmost, middle and rightmost charges with labels L, M and R. •
The field E L due to the leftmost charge has positive x and y components.
•
The field E R due to the rightmost charge has a negative x component and a positive y component.
• The field E M due to the middle charge has no x component and a negative y component. • † When these three fields are added, the x components of E L and E R will cancel out, leaving only their y components to be added to the y component of E M . Therefore, we will only calculate the y † component of the field E L and algebraically add twice this value to the negative y component of † E M . Hereafter, we calculate the magnitudes of the fields. † † 2 ˆÊ -6 ˆ QL Ê N ⋅ m 10 ¥10 C ˜ = 3.6 † EL = ke 2 = Á 9 ¥109 ¥10 3 N/C ˜Á rLA Ë † C2 ¯ÁË ( 5.0 m)2 ˜¯ Ê3 mˆ 3 3 ELy = EL sin q = ( 3.6 ¥10 3 N/C)Á ˜ = 2.16 ¥10 N/C . Similarly, ERy = 2.16 ¥10 N/C . Ë5 m¯ Ê Q N ⋅ m2 ˆÊ 6.0 ¥10-6 C ˆ ˜ = 6.0 ¥10 3 N/C , along the negative y axis. EM = ke 2M = Á9 ¥109 ˜Á rMA Ë C 2 ¯ÁË ( 3.0 m)2 ˜¯ † EMy = -6.0 ¥10 3 N/C .
†
† †
The y-component of the field due to all three charges is
† †
E y = ELy + ERy + EMy = 2.16 ¥10 3 + 2.16 ¥10 3 + (-6.0 ¥10 3 ) = -1.68 ¥10 3 N/C .
The negative sign of the y-component emphasizes what was argued above, viz., that the electric field at A is directed downwards, towards the negative y axis. It can be written in vector notation as †
E = -1.68 ¥10 3 ˆj N/C .
†
Name
Solutions
2. (a) Calculate the electric potential at the point A. VA = ke
12 points
QL Q Q Ê N ⋅ m2 ˆÊ10 ¥10 -6 C -6.0 ¥10-6 C 10 ¥10 -6 C ˆ + ke M + ke R = Á 9 ¥109 + + ˜Á ˜ rLA rMA rRA Ë C2 ¯Ë 5.0 m 3.0 m 5.0 m ¯
Ê N ⋅ m2 ˆ -6 -6 -6 = Á 9 ¥10 9 ˜(2 ¥10 C/m - 2 ¥10 C/m + 2 ¥10 C/m) 2 C ¯ Ë
†
Ê N ⋅ m2 ˆÊ -6 C ˆ 4 4 4 = Á 9 ¥10 9 ˜ = 1.8 ¥10 N ⋅ m/C = 1.8 ¥10 J/C = 1.8 ¥10 V ˜Á2 ¥10 2 C ¯Ë m¯ Ë
†
†
(b) If the electric potential at the point B is zero, what is the ratio of the distances y and s? 8 points
Ê10 ¥10-6 6.0 ¥10 -6 10 ¥10 -6 ˆ Ê QL QM QR ˆ QL QM QR VB = ke + ke + ke = ke Á + + ˜ = ke Á + ˜ rLB rMB rRB y s ¯ s y s Ë s Ë ¯ Ê 20 ¥10 -6 6.0 ¥10-6 ˆ = ke Á ˜. s y Ë ¯
†
VB = 0
fi
†
20 ¥10-6 6.0 ¥10-6 = 0. s y
The last equation can be rearranged to give the result
†
†
y = 0.3s , or
y / s = 0.3 , or s = 10y / 3, or s / y = 10 / 3 = 3.33 . 3. Calculate the electric potential energy of this assembly of charges.
U = ke
QLQR QQ Q Q + ke† L M + ke M R rLA rLM rMR
†
20 points
†
Ê (10 ¥10-6 C)(10 ¥10-6 C) (10 ¥10 -6 C)(-6.0 ¥10 -6 C) (-6.0 ¥10-6 C)(10 ¥10-6 C) ˆ ˜ = ke ÁÁ + + ˜ 8.0 m 4.0 m 4.0 m Ë ¯
†
†
Ê N ⋅ m2 ˆ -12 2 -12 2 -12 2 = Á 9 ¥10 9 ˜(12.5 ¥10 C /m -15.0 ¥10 C /m -15.0 ¥10 C /m) 2 C ¯ Ë
†
Ê N ⋅ m2 ˆ -12 2 = Á 9 ¥10 9 ˜(-17.5 ¥10 C /m) = -0.1575 J 2 C ¯ Ë
†
Name
Solutions
4. (a) In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a
proton, where the radius of the orbit is 5.29 ¥10 -11 m. Calculate the electric potential due to the proton at the position of the electron. 8 points V = ke
Q Ê N ⋅ m2 ˆÊ +1.60 ¥10-19 C ˆ = Á9 ¥109 ˜Á † ˜ = 27.2 V r Ë C2 ¯Ë 5.29 ¥10-11 m ¯
†
(b) Calculate the electric potential energy of the electron. State your answer in joules as well as in electron-volts. 6 points U = qelectronV = (-e)(27.2 V) = -27.2 eV
†
The same potential energy can be expressed in joules by inserting the numerical value of the electron charge.
U = qelectronV = (-1.6 ¥10-19 C)( 27.2 V) = -4.35 ¥10-18 J
† (c) If the electron moves at a speed of 2.18 ¥106 m/s , calculate its kinetic energy. State your answer in joules as well as in electron-volts. (Ignore relativistic effects.) 4 points K = 21 melectron v2 =
†
1 2
(9.11¥10 †
2
kg)(2.18 ¥106 ) = 2.16 ¥10 -18 J
The same kinetic energy can be expressed in electron-volt units by using the conversion factor 1 eV = 1.6 ¥10-19 J .
K = 2.165 ¥10 -18 J ¥ †
-31
1 eV = 13.53 eV 1.60 ¥10 -19 J
(d) What is the total energy of the electron? State your answer in electron-volts.
†
Total energy is the sum of kinetic and potential energies. E = 13.53 eV+ (-27.2 eV) = -13.67 eV
†
2 points
Name
Solutions
5. (a) A parallel-plate capacitor is made by depositing metal coatings on both faces of a flat,
circular disk of fused quartz. The disk is 1.00 mm thick and has a diameter of 27.6 cm. Dielectric constant of fused quartz is 3.78. Calculate the capacitance of this capacitor. 10 points 2
pd 2 p (0.276 m) Area of plates, A = = = 6.0 ¥10-2 m2 . 4 4
(8.85 ¥10-12 F/m)(6.0 ¥10-2 m2 ) = 2.0 ¥10-9 F . eA Capacitance, C = k 0 = 3.78 ¥ d 1.00 ¥10-3 m † †
(b) What potential difference should be maintained between the two “plates” in order to store 25.0!mJ of energy in the capacitor? 6 points When a potential difference V is maintained between the plates, the energy stored in the capacitor is
U = 21 CV 2 . Solving this for the unknown V, we get
† V=
2(25.0 ¥10-3 J) 2U = = 5000 V . C 2.0 ¥10-9 F
(c) What will be the magnitude of the charge on each plate under the conditions of part (b)? 4 points
†
Q = CV = ( 2.0 ¥10 -9 F)(5000 V) = 1.0 ¥10-5 C = 10 mC
†
General Physics II
Solutions Spring 2003 Dr. Satinder S. Sidhu
Test 3 Questions 1 through 3 all refer to the situation in the following figure. It shows three charges placed on the horizontal axis and two empty points A and B on the vertical axis. Unless directed otherwise, you should include the effects of all three charges in all calculations.
1. Calculate the electric field at the point A. Clearly state its magnitude and direction.
20 points
In the following, I refer to the leftmost, middle and rightmost charges with labels L, M and R. •
The field E L due to the leftmost charge has positive x and y components.
•
The field E R due to the rightmost charge has a negative x component and a positive y component.
• The field E M due to the middle charge has no x component and a negative y component. • † When these three fields are added, the x components of E L and E R will cancel out, leaving only their y components to be added to the y component of E M . Therefore, we will only calculate the y † component of the field E L and algebraically add twice this value to the negative y component of † E M . Hereafter, we calculate the magnitudes of the fields. † † 2 ˆÊ -6 ˆ QL Ê N ⋅ m 10 ¥10 C ˜ = 3.6 † EL = ke 2 = Á 9 ¥109 ¥10 3 N/C ˜Á rLA Ë † C2 ¯ÁË ( 5.0 m)2 ˜¯ Ê3 mˆ 3 3 ELy = EL sin q = ( 3.6 ¥10 3 N/C)Á ˜ = 2.16 ¥10 N/C . Similarly, ERy = 2.16 ¥10 N/C . Ë5 m¯ Ê Q N ⋅ m2 ˆÊ 6.0 ¥10-6 C ˆ ˜ = 6.0 ¥10 3 N/C , along the negative y axis. EM = ke 2M = Á9 ¥109 ˜Á rMA Ë C 2 ¯ÁË ( 3.0 m)2 ˜¯ † EMy = -6.0 ¥10 3 N/C .
†
† †
The y-component of the field due to all three charges is
† †
E y = ELy + ERy + EMy = 2.16 ¥10 3 + 2.16 ¥10 3 + (-6.0 ¥10 3 ) = -1.68 ¥10 3 N/C .
The negative sign of the y-component emphasizes what was argued above, viz., that the electric field at A is directed downwards, towards the negative y axis. It can be written in vector notation as †
E = -1.68 ¥10 3 ˆj N/C .
†
Name
Solutions
2. (a) Calculate the electric potential at the point A. VA = ke
12 points
QL Q Q Ê N ⋅ m2 ˆÊ10 ¥10 -6 C -6.0 ¥10-6 C 10 ¥10 -6 C ˆ + ke M + ke R = Á 9 ¥109 + + ˜Á ˜ rLA rMA rRA Ë C2 ¯Ë 5.0 m 3.0 m 5.0 m ¯
Ê N ⋅ m2 ˆ -6 -6 -6 = Á 9 ¥10 9 ˜(2 ¥10 C/m - 2 ¥10 C/m + 2 ¥10 C/m) 2 C ¯ Ë
†
Ê N ⋅ m2 ˆÊ -6 C ˆ 4 4 4 = Á 9 ¥10 9 ˜ = 1.8 ¥10 N ⋅ m/C = 1.8 ¥10 J/C = 1.8 ¥10 V ˜Á2 ¥10 2 C ¯Ë m¯ Ë
†
†
(b) If the electric potential at the point B is zero, what is the ratio of the distances y and s? 8 points
Ê10 ¥10-6 6.0 ¥10 -6 10 ¥10 -6 ˆ Ê QL QM QR ˆ QL QM QR VB = ke + ke + ke = ke Á + + ˜ = ke Á + ˜ rLB rMB rRB y s ¯ s y s Ë s Ë ¯ Ê 20 ¥10 -6 6.0 ¥10-6 ˆ = ke Á ˜. s y Ë ¯
†
VB = 0
fi
†
20 ¥10-6 6.0 ¥10-6 = 0. s y
The last equation can be rearranged to give the result
†
†
y = 0.3s , or
y / s = 0.3 , or s = 10y / 3, or s / y = 10 / 3 = 3.33 . 3. Calculate the electric potential energy of this assembly of charges.
U = ke
QLQR QQ Q Q + ke† L M + ke M R rLA rLM rMR
†
20 points
†
Ê (10 ¥10-6 C)(10 ¥10-6 C) (10 ¥10 -6 C)(-6.0 ¥10 -6 C) (-6.0 ¥10-6 C)(10 ¥10-6 C) ˆ ˜ = ke ÁÁ + + ˜ 8.0 m 4.0 m 4.0 m Ë ¯
†
†
Ê N ⋅ m2 ˆ -12 2 -12 2 -12 2 = Á 9 ¥10 9 ˜(12.5 ¥10 C /m -15.0 ¥10 C /m -15.0 ¥10 C /m) 2 C ¯ Ë
†
Ê N ⋅ m2 ˆ -12 2 = Á 9 ¥10 9 ˜(-17.5 ¥10 C /m) = -0.1575 J 2 C ¯ Ë
†
Name
Solutions
4. (a) In the Bohr theory of the hydrogen atom, an electron moves in a circular orbit about a
proton, where the radius of the orbit is 5.29 ¥10 -11 m. Calculate the electric potential due to the proton at the position of the electron. 8 points V = ke
Q Ê N ⋅ m2 ˆÊ +1.60 ¥10-19 C ˆ = Á9 ¥109 ˜Á † ˜ = 27.2 V r Ë C2 ¯Ë 5.29 ¥10-11 m ¯
†
(b) Calculate the electric potential energy of the electron. State your answer in joules as well as in electron-volts. 6 points U = qelectronV = (-e)(27.2 V) = -27.2 eV
†
The same potential energy can be expressed in joules by inserting the numerical value of the electron charge.
U = qelectronV = (-1.6 ¥10-19 C)( 27.2 V) = -4.35 ¥10-18 J
† (c) If the electron moves at a speed of 2.18 ¥106 m/s , calculate its kinetic energy. State your answer in joules as well as in electron-volts. (Ignore relativistic effects.) 4 points K = 21 melectron v2 =
†
1 2
(9.11¥10 †
2
kg)(2.18 ¥106 ) = 2.16 ¥10 -18 J
The same kinetic energy can be expressed in electron-volt units by using the conversion factor 1 eV = 1.6 ¥10-19 J .
K = 2.165 ¥10 -18 J ¥ †
-31
1 eV = 13.53 eV 1.60 ¥10 -19 J
(d) What is the total energy of the electron? State your answer in electron-volts.
†
Total energy is the sum of kinetic and potential energies. E = 13.53 eV+ (-27.2 eV) = -13.67 eV
†
2 points
Name
Solutions
5. (a) A parallel-plate capacitor is made by depositing metal coatings on both faces of a flat,
circular disk of fused quartz. The disk is 1.00 mm thick and has a diameter of 27.6 cm. Dielectric constant of fused quartz is 3.78. Calculate the capacitance of this capacitor. 10 points 2
pd 2 p (0.276 m) Area of plates, A = = = 6.0 ¥10-2 m2 . 4 4
(8.85 ¥10-12 F/m)(6.0 ¥10-2 m2 ) = 2.0 ¥10-9 F . eA Capacitance, C = k 0 = 3.78 ¥ d 1.00 ¥10-3 m † †
(b) What potential difference should be maintained between the two “plates” in order to store 25.0!mJ of energy in the capacitor? 6 points When a potential difference V is maintained between the plates, the energy stored in the capacitor is
U = 21 CV 2 . Solving this for the unknown V, we get
† V=
2(25.0 ¥10-3 J) 2U = = 5000 V . C 2.0 ¥10-9 F
(c) What will be the magnitude of the charge on each plate under the conditions of part (b)? 4 points
†
Q = CV = ( 2.0 ¥10 -9 F)(5000 V) = 1.0 ¥10-5 C = 10 mC
†
