Phys 05, First Test. September 0, 011 50 minutes Name: Show all work for maximum credit. Each problem is worth 10 points. Work 10 of the 11 problems. k = 9.0 x 10 9 N m / C ε 0 = 8.85 x 10-1 C / N m e = 1.60 x 10-19 C ρ = 1.68 x 10-8 Ω m for copper 1) Given the two charges shown in the figure, at what position x is the electric field zero? The two charges are separated by a distance of meters. 9 Q -4 Q m For the electric field to be zero: E 9Q + E 4Q = 0, so E 9Q = - E 4Q 9Q 4Q 9 4 3 k k 3x 4 x x 4 m ( x) x ( x) x x x ) Find the electric potential at the origin of coordinates due to the charges located at and B. Consider the coordinates given in meters and charges Q = 4C and Q B = -5C. 6 Q 9 410 V k 910 1,000 volt r 3 6 QB 9 ( 510 ) VB k 910 -,500 volt r So the net electric potential is: V T = V + V B = 1000 + (-500) = -10,500 V 3) Determine the electric field at the origin of coordinates due to the charges located at and B using the figure in the previous problem. Give your answer as a magnitude and an angle with the proper units.
First let s find the magnitudes of the two vectors using the formula for the electric field of a point charge: 6 E q Nm 4x10 C k 9 9x10 4,000 N/C down since positive charges repel r C (3m) 6 E q Nm 5x10 C k B 9 B 9x10 11,50 N/C to the right since opposite charges attract r C (m) Now we consider the directions of these vectors: E : Since q is positive the electric field vector will point away from the charge, so it will be a vertically downward vector. (0,-4000) E B : Since q B is negative the electric field vector will point toward the charge, so it will be a horizontal vector pointing to the right. (11,50,0) We can add these vectors easily: E + E B = (11,50, -4000) Ex Ey 11,50 4,000 E 11,900 N/C The angle is computed: θ = tan -1 E ( ) = -0 o. E B 4) capacitor is made with two plates submerged in water (K water =81). If their area is 3.11 m and the distance between them is 0.66 mm. Calculate the capacitance (in Farads) and how much charge is stored in the capacitor when it is connected to a voltage of 10 volts. 1 3.11m C K o (81)(8.8510 F / m) 3.4 μf d 0. 00066m nd the charge stored is: Q CV 3.4F(10V) 410 μc
5) Find the power delivered by the voltage source. The voltage is given in RMS voltage. lso, find the current through resistor R 1. To find the power delivered by the voltage source, notice that the three resistors are equivalent of one. To find the equivalent, we see that the 1 and the 4 resistors are in 1 parallel, so they are equivalent to one of value R equivalent 8, and then 1 1 1 4 this equivalent resistor is in series with the 10 resistor, giving a total of 18. So the power is: (90V ) Power 450W 18 To find the current through R 1, notice that the total current through the whole circuit is the same as the current through R 1. We don t know the voltage across R 1., but we do know the voltage across the whole circuit. We can then use Ohm s Law. V 90V I 5.0. R 18 eq 6) The switch in the figure has been closed for a long time, so the capacitor is charged to 30 volts. Then at time t = 0 you open the switch and the voltage starts dropping (the multimeter behaves as a resistance of 1.5MΩ) find how long it takes for the voltage in the capacitor to get to 5V. The time constant is RC ( 1.5M)(18F ) 7seconds The equation for discharge is: V t / c V e
/ 470 With the values of the problem: 5 30e t Solving for t we get: 5 / 5 5 e t 7 ln t / 7 t 7ln 48 seconds 30 30 30 7.) Determine the capacitance of the following arrangement of capacitors and find how much energy is stored when you apply 4V between terminals and B. To determine the capacitance of this arrangement of capacitors notice that the branch on the left side is made of 3 capacitors in series. That branch is equivalent to one capacitor of value: 1 C 6F 1 1 1 18F 18F 18F Then, this equivalent capacitor is in parallel with an 18µF capacitor giving a total of 4µF together. nd the energy stored is: 1 1 E CV 4nF(4V) 6.9 mj 8.) point charge of 8.50 nc is located at the origin and a second charge of -3.10 nc is located on the X-axis at X = 3.00 m. Calculate the electric flux through a sphere centered at the origin with a radius of 1.50 m. Only the 8.50 nc charge is enclosed by the sphere. Q 8.50nC Ed 960.5 N m / C 1 0 8.85x10 9.) We want to find the current through the resistors R 1, R and R 3. Draw the direction of the currents on the diagram. Write down the equations that you need to solve the problem. You will get full credit if the equations are correct (signs, quantities, etc.) You do not need to solve the equations.
+ - - + + I 1 I 3 - I + - The node equation is: I = I 1 + I 3 The loop equations are: 10 35I 1 45I 0 left loop (I chose clockwise) 5I 45I 15I 0 right loop (I chose clockwise) 5 3 3 nd writing them in the standard form: - I 1 + I - I 3 = 0 35I 1 45I... 10 45I 0I 3 5 10) The ring shown in the figure has a uniform charge Q and radius R. Determine the electric field at point P, which is located on the axis of the ring a distance x from its center. To calculate the electric field notice that we can divide the ring into differentials of charge. Each one contributing an electric field equal to: dq dq de k k r x R
But notice that only the component of the vector in the vertical direction will contribute to the integral. The other components will give zero due to the symmetry of the problem, so: dq dq x de x k cos k x R x R x R nd after integrating: Qx E k x R 3/ 11) wire is uniformly tapered from a radius R 1 = 1.0 mm to a final radius R = 3.0 mm. The length of the wire is 0 m, so the equation that describes the radius is: r 0.001 0. 0001x Calculate the resistance of the wire [The resistivity of copper is 1.68 x 10-8 Ω m] L R 4, where a and b are the diameters at the two ends. ab 8 0 R 4(1.6810 ) 0.036 Ω 3.1416(0.00)(0.006) Note: If you don t remember the equation above you can always deduce it by integration: R L dx as we did in class. b a a x L 0