Physics 212 Test III March 7, 2018 Write all answers on the blank sheets. Work w

Question

Physics 212 Test III March 7, 2018 Write all answers on the blank sheets. Work written on the test itself will ot be grad Assume that all values are given to three significant figures 1. (15 points) As shown in the picture, three charges are fixed on three comers of a square of side length 10.0 cm. a. Find the electric potential at the center of the square. b. Find the potential at the coner not occupied by a charge. c. Compute the work an external agent would have to do to move an electron from the center to the unoccupied corner. 2. (15 points) a. A capacitor is charged up by connecting it across a 25.0 v battery. The resulting energy stored is found to be 1.20 . Find the value of the capacitance, and the magnitude of the charge on the plates. b. A capacitor is constructed from two large square sheets of tin foil; each square has a side length of 40.0 cm. The tin foil sheets are separated by a large piece of construction paper having a thickness of 0.70 mm and a dielectric constant of 1.70. Find the capacitance. 3. (10 points) Show how to convert a galvanometer into a voltmeter that reads 10.0 v at full scale. The galvanometer has an internal resistance of 35.0 , and deflects full scale with a current of 2.00 mA. Include a sketch showing in detail how the voltmeter is to be constructed. lbv, 12 18 4. (12 points) For the circuit shown at the right, write down a sufficient number of equations so that the designated currents can be solved for. You do not have to solve the equations. 15.2 20V

Explanation / Answer

1) a) potential due to a point charge q at a distance d is V = ke*q/d

the length of diagonal is sqrt(2)*10 = 14.14 cm = 0.1414 m

the distance from every charge and to the center is r = 0.1414/2 = 0.0707 m

V_center = V1+V2 + V3

V_center = (ke*q1/r) + (ke*q2/r)+(ke*q3/r)

V_center = (ke/r)(q1+q2+q3)

V_center = (9*10^9/0.0707)*(3+2-4)*10^-6 = 1.27*10^5 volts

b) V_unoccupied = V1+V2+V3

V_unoccupied = (ke*q1/r1)+(ke*q2/r2)+(ke*q3/r3)

V_unoccupied = ke*((q1/r1)+(q2/r2)+(q3/r3))

V_unoccupied = 9*10^9*((3*10^-6/0.1)+(2*10^-6/0.1414)-(4*10^-6/0.1)

V_unoccupied = 0.373*10^5 volts

then

C) work is w = q*(V_unouupied - V_center)

W = 1.6*10^-19*(0.373-1.27)*10^5 = -1.45*10^-14 J

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