# Exam 1 Solutions. PHY2054 Fall Prof. Paul Avery Prof. Andrey Korytov Sep. 26, 2014

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1 Exam 1 Solutions Prof. Paul Avery Prof. Andrey Korytov Sep. 26, Charges are arranged on an equilateral triangle of side 5 cm as shown in the diagram. Given that q 1 = 5 µc and q 2 = q 3 = 2 µc find the magnitude of the net force on charge q 1 (in N). Answer: 62 Solution: The x component of the force cancels and the y components of the q 2 q 3 force on q 1 are the same for both charges. Let a be the length of each side. Then the total force in the y direction is 2 q 1 q 2 a 2 k sin60 = 62 N. q 1 2. Charges are arranged on a square of side d as shown in the diagram. In what direction does the electric field at the center of the square point? (The quadrants are numbered counterclockwise starting from the positive x-axis.) Answer: Fourth quadrant Solution: The electric fields at the origin due to the left & right (+Q and Q) and top & bottom (+Q and 2Q) charges do not cancel. The electric fields due to these two charge pairs are then in the directions +x (for the left & right charges) and y (for the top & bottom charges), leading to a net electric field in the 4 th quadrant. 3. Two charged particles are fixed to the x-axis: particle 1 of charge q 1 = 20 µc at x = 0 m, and particle 2 of charge q 2 = 80 µc at x = 0.6 m. At what coordinate along the x-axis is the net electric field produced by the particles equal to zero? Answer: 0.6 m Solution: Between x = 0 and x = 0.6 m, electric fields due to charges q 1 and q 2 point in the same direction and cannot cancel. For x > 0, the two fields are in opposite directions, but the larger in magnitude charge q 2 is closer and hence its field is always greater than the field due to charge q 1 ; hence, the two fields cannot cancel each other here either. For x < 0, the fields are again in the opposite directions and would cancel, if their magnitudes are the same: Solving yields x = L = 0.6 m. q 1 x k = q 2 2 ( L x) k. 2 1

2 4. An electron with velocity v 0 along the +x direction enters a region with a uniform electric field of magnitude 5000 V/m in the y direction. If the electron travels 5.0 cm in the x direction and gets deflected by 1.0 cm in the y direction what is v 0 (in m/s)? Answer: Solution: The deflection by a constant acceleration is y = 1 2 at2, where y = 0.05 m, the acceleration a = Ee / m e and the time t = Δx / v 0. Solving yields v 0 = m/s. 5. A conducting sphere with charge +10 nc is placed at the center of two concentric conducting spherical shells of radius r 1 = 2 cm and r 2 = 5 cm (measured to the outer surfaces). The inner shell carries a charge of 7 nc and the outer shell has a charge of +6 nc. Find the charge (in nc) on the inner surface of the outer shell. Answer: 3 Solution: The E field inside the conductor of the outer shell must be 0. A Gaussian surface drawn within the outer shell will thus have no flux going through it, implying, from Gauss law, that there is 0 charge enclosed by the Gaussian surface. Since the total charge of the central sphere and inner shell is +3 nc, the charge on the inner surface of the outer shell must therefore be 3 nc. 6. In the previous problem, find the magnitude of the electric field at r = 50 cm. Answer: 324 V/m Solution: Outside the uniformly charged shells/spheres, the electric field is kq / r 2 and points radially. In this problem, the total field due to the two shells and the central sphere is then where Q = +9 nc. This yields E = 324 V/m/ E = kq 1 / r 2 + kq 2 / r 2 + kq 3 / r 2 = q 1 + q 2 + q 3 r 2 k = Q r 2 k, 2

3 7. Four charges of magnitude Q = 3.0 µc (but different signs, as in the figure) are arranged on the corners of a square of side 25 cm. Find the potential energy of the system of the four charges (in J). Answer: 0.84 Solution: For a system of point charges, the total potential energy is U = all (i,j) pairs kq i q j r With four charges, there are 6 pairs: 12, 13, 14, 23, 24, 34. Let L be the length of a side. Four pairs (corresponding to the 4 sides) each contribute kq 2 / L while the diagonals each contribute kq 2 / 2L. The total potential energy is thus +Q -Q -Q +Q 4 kq2 + 2 kq2 = kq2 L 2L L ( ) = 0.84 J. 8. Electrons in a particle beam have a kinetic energy of J. What is the magnitude of the electric field (in V/m) that will stop these electrons in a distance of 0.1 m? Answer: 2000 Solution: The distance d can be calculated from K = Fd = Eed, yielding E = 2000 V/m. 9. The movement of a charge in an electric field from one point to another at constant speed without the expenditure of work by or against the field Answer: none of these Solution: The work is zero only when moving along an equipotential surface. 10. Two particle each with charge Q are fixed at the vertices of an equilateral triangle with sides of length a. The work required to move a particle with a charge q from the other vertex to the center of the line joining the fixed charges is Answer: 2kQq / a Solution: The work required of an external force to move the charge q equals to the change in the potential energy of that charge:! Qq W ext = ΔU =U f U i = 2# " (0.5a) k \$! & 2# Qq % " a k \$ & = 2 Qq % a k. a a a 11. Two isolated conducting spheres are separated by a large distance. Sphere 1 has a radius of R and an initial charge 3Q while sphere 2 has a radius of 3R and an initial charge 7Q. A very thin 3

4 copper wire is now connected to the spheres to allow charge to flow between the spheres. How much charge will be transferred from sphere 2 to sphere 1? (Note that the charge transferred can be positive, negative or zero.) Answer: Q / 2 Solution: After reaching equilibrium, the two spheres must be at the same potential, thus Q 1 R k = Q 2 3R k Let ΔQ be the charge taken from sphere 2 and moved to sphere 1. Then: from where we obtain ΔQ = Q / 2. 3Q + ΔQ R k = 7Q ΔQ 3R k 12. Four protons are placed at rest on the vertices of a square of side 2.0 µm. The protons are released simultaneously. When the protons are very far away, what is their speed (in m/s)? Answer: 430 Solution: Conservation of total energy yields 0 +U i = K f + 0, where the initial potential energy ( )( 4 + 2) (see problem 7) and the final kinetic energy of four protons ( ). From here, we obtain v = ( ke2 / ml) ( 2 +1/ 2) = 432 m/s. U i = ke 2 / L K = mv2 13. Two equipotential surfaces lying near the middle of the space between the plates of a parallel-plate capacitor are 2.0 mm apart and have a potential difference of volt. The area of each plate is 7.5 cm 2. What is the magnitude of the charge on each plate, (in units of C)? Answer: 4 Solution: The electric field between two uniformly charged plates with charges Q and Q is E = σ / ε 0 = Q / ε 0 A, where A is the area of each plate. We calculate E = ΔV / d = 0.6 V/m. Solving for Q yields Q = C. 4

5 14. A certain parallel plate capacitor with capacitance 12 µf is connected to a source of EMF with potential 3 V. A dielectric material with κ = 4 is then inserted between the plates of the capacitor with the capacitor still connected to the circuit. By how much does the energy stored in the capacitor change? Answer: J Solution: The energy in a capacitor is U = 1 2 CV 2, with V the voltage across it. Since the capacitor remains connected to the emf source when the dielectric is inserted, the change in energy is caused only by the change in capacitance ΔC = C f C i = κc C = κ 1 ( )C. Thus the change in energy is ΔU = ( ) 3 2 = 162µJ. 15. An air-filled parallel-plate capacitor has a capacitance of 2 pf. The plate separation is then doubled and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 4 pf. The dielectric constant of the wax is: Answer: 4.0 Solution: The capacitance of a parallel-plate capacitor is C = κε 0 A d Without the dielectric, the doubling of the plate spacing would change the capacitance from 2 pf to 1 pf. Thus to achieve 4 pf, the dielectric constant must be A parallel plate capacitor with a capacitance of 2.0 nf is charged to have 0.8 µc on each plate. How much work must be done by an external agent to double the plate separation while keeping the charge constant? Answer: J Solution: If the plate separation is doubled, C new = C 0 / 2. With the charge unchanged, the change in energy is thus ΔU = Q2 2C new Q2 2C 0 = Q2 2C 0 = +160µJ. 5

6 17. A 9 V battery is connected to a 3 Ω resistor. How much charge passes through the resistor in 3 hours? Answer: 32,400 C Solution: The total charge Q = i t, where i = 3 A and t = 10,800 s. Thus Q = 32,400 C. 18. You have two resistors R 1 and R 2, made from the same material, with the length and diameter of R 1 both double that of R 2. If R 1 and R 2 are connected in parallel to a battery, find the ratio of the currents running through R 1 and R 2 (find I 1 / I 2 ). Answer: 2 Solution: The resistance of a cylindrical resistor is R = ρl A = ρl. The cross sectional area 2 π (d / 2) of R 1 is 4 times that of R 2 while its length is double, therefore R 1 = 1 2 R 2. The current in R 1 is thus twice the current in R In the circuit diagram shown find the voltage drop across the 3.0 Ω resistor. Answer: 4.5 V Solution: There is 12 V applied across the bottom branch, whose total 3.0 Ω resistance is = 8 Ω (resistors connected in series). Thus the current is 12 V / 8 Ω = 1.5 A and the voltage drop across the 3 Ω resistor is 1.5 A 3 Ω = 4.5 V. 12 V 6.0 Ω 12 Ω 4.0 Ω 5.0 Ω 20. In the circuit shown find the total energy (in µj) stored on all the capacitors. Where C 1 = 2 µf, C 2 = 4 µf, C 3 = 9 µf, C 4 = 1 µf, C 5 = 3 µf, and V = 10V. Answer: 182 Solution: The total energy can be calculated using the equivalent capacitance and the emf since from the battery s perspective the equivalent capacitance behaves in exactly the same way as the original capacitors. C 4 and C 5 combine to give C 45 = 3.0 µf. Then C 2, C 45, C 3 in series give 1.64 µf. Adding in C 1 in parallel then gives a total capacitance of 3.64 µf. The energy is then U = 1 2 CV 2 = = 182µJ. V C 1 C 2 C 3 C 4 C 5 6

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