1. How many coulombs of positive charge are contained in the atomic nuclei of this gas? You have 5.0 mol of O2 gas. Using Avogadro’s Number, first we calculateSo, O2 gas is 8 protons in a.0 mol • 010000000000000000000000 par ticles51 mol6.02e23 particles= 3 single atom. Oxygen is diatomic though so O2 gas will have 16 protons.010000000000000000000000 • 48160000000000000000000000 pr otons31 particle16 protons= or 7.715232e+15 nC.8160000000000000000000000 • 715232 Coulombs 41 proton1.602e−19 Coulombs= 7 2. A plastic rod that has been charged to -15.0 nC, touches a metal sphere. Afterward, the rod's charge is -10.0 nC. What kind of charged particle was transferred between the rod and the sphere? An electron. In which direction was the charged particle transferred between rod and sphere? That is, did it move from the rod to the sphere or from the sphere to the rod? It moved from rod to sphere. How many charged particles were transferred? -15.0 nC + 5 nC (transferred) = -10.0 nC nC • 0.000000005 Coulombs51e−9 Coulombs1 nano Coulombs= .000000005 Coulombs • 1210986267.1660424469413233458177278 electrons 01 electrons 1.602e−19 Coulombs= 33.125 e 10 3. A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? (e = 1.60 × 10-19 C) 1.25e13 4. Electrically neutral objects cannot exert an electrical force on each other, but they can exert a gravitational force on each other. False 5. A 30 nC charge experiences a 0.032 N electric force. What is the magnitude of electric field at the position of this charge? (0.032/30)*10^9 = 1066666.6666666666666667 6. A protein molecule in an electrophoresis gel has a negative charge. The exact charge depends on the pH of the solution, but 30 excess electrons is typical. What is the magnitude of the electric force on a protein with this charge in a 1700 N/C electric field? F = Eq = 1700N/C*(30*1.60x10^-19 C) = 0.00000000000000816 7. Large electric fields in cell membranes cause ions to move through the cell wall. The field strength in a typical membrane is 1.0×107N/C. What is the magnitude of the force on a calcium ion with charge +e? Required Force = 1.0*(10⁷)*1.6*10⁻¹⁹ = 1.6*10⁻¹² N 8. A proton is placed in an electric field of intensity 800 N/C. What are the magnitude and direction of the acceleration of the proton due to this field? (e= 1.60 × 10-19 C, mproton = 1.67 × 10-27 kg) 7.66 × 1010 m/s2 in the direction of the electric field 9. A particle with a charge of +4.0 μC has a mass of 5.0 g. What magnitude electric field directed upward will exactly balance the weight of the particle? 1.2 × 104 N/C 10. A 10 nC charge sits at a point in space where the magnitude of the electric field is 1300 N/C .What will the magnitude of the field be if the 10 nC charge is replaced by a 20 nC charge? Assume the system is big enough to consider the charges as small test charges. 1300N/C 11. What magnitude charge creates a 1.0 N/C electric field at a point 1.0 m away? E=K x q /d^2 then K=9 x 10^9 N.m^2 /C^2 then E=1.0 then q= E x d^2 / K then = (1 x 1^2)/ (9 x 10^9) and finally = 0.11 x 10^-9 C = 0.1112. What is the strength of the electric field 3.0 cm from a small glass bead that has been charged to + 5.0 nC? It is E = F/Q = kQ/r² then (8.99e9)(5e-9) / (0.03)² = 49944.44 N/C and then What is the direction of the electric field 3.0 cm from a small glass bead that has been charged to 5.0 nC ? away from the bead. 13. What is the strength of the electric field Ep 7.0 mm from a proton? E = kQ/r² .. (N/C or V/m) then E = (9.0E9)(1.60E-19) / (7.0E-3)² ►= 9.0E-5 N/C (direction: radially outward from p) then 0.000029387755102040816327 is the answer. What is the direction of the electric field Ep 7.0 mm from a proton? Away from the proton. What is the strength of the electric field Ee 7.0 mm from an electron? 0.000029387755102040816327 but What is the direction of the electric field Ee 7.0 mm from an electron? Toward the electron 14. A + 19 nC charge is located at the origin. What is the strength of the electric field at the position (x,y)=(5.0cm,0cm)? E = kq/r² = kq/(x²+y²) then What is the strength of the electric field at the position (x,y)=(−5.0cm,5.0cm)? 8.988e9*19e-9/(-0.05² + 0.05²) = 34,153 N/C then What is the strength of the electric field at the position (x,y)=(−5.0cm,−5.0cm)? 8.988e9*19e-9/(-0.05² + -0.05²) = 34,153
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