PHYS 208 Final: PHYS 208 TAMU Final 2010 Solutions

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turquoisegnu128

31 Jan 2019

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The circuit below was put together a long time ago so that the steady state has been reached. Find all currents, voltage drops across the resistors, and the charge on the capacitor plates. i-i. There is a charge q uniformly spread over a quarter of circle of radius r centered at the origin. There is also a point positive charge q0 located at the origin. Find a force acting on charge q0. y dq x. In the circuit below the switch has been closed for a long time. : ---=o: if at t0 the switch is opened, what differential equation can be solved to find q, the the capacitor, as a function of time? (assume that l includes the self-inductance). ,2q: find the charge on the capacitor plates as a function of time if inductance l can be ignored. _: find the current through r2 as a function of time if capacitance c can be ignored.

Related Questions

A charge q that moves between two points A and B, and has anelectric force acting on it ( and perhaps several other forces) haswork W done on it by the electric force. The electrostatic force isconservative and so we can define a change in electric potentialenergy, Delta U = -W. We defined electric field using the relationF^^\-> = q E^^\->, and in a similar way we define electricpotential commonly called voltage as
Delta V = V_B - V_A = -W/q=(Delta U)/q or equivalently q Delta V =q (V_B - V_A) = -W=Delta U

A ball has mass 3.0 Âµg and charge -6 nC moves from point A to pointB. It has the same speed at points A and B, and the work done bythe electric field during the move is -1.5 mJ.
(a) What is the change in electric potential energy of the ball?1
(b) What is the change in voltage between points A and B? 2
(c) At which point does the ball have a higher electric potentialenergy?
(d) Which point has the higher voltage?
(e) Has the kinetic energy of the ball changed?

A circuit is constructed with four resistors, one capacitor,one battery and a switch as shown. The values for the resistorsare: R1 = R2 = 65 O, R3 = 115 O and R4 = 136 O. The capacitance isC = 79 ÂµF and the battery voltage is V = 24 V.

1) The switch has been open for a long timewhen at time t = 0, the switch is closed. What is I1(0), themagnitude of the current through the resistor R1 just after theswitch is closed?

___A

2) What is I1(8), the magnitude of the current thatflows through the resistor R1 a very long time after the switch hasbeen closed?

___A

3) What is Q(8), the charge on the capacitorafter the switch has been closed for a very long time?

___ÂµC

4) What is I2(8), the magnitude of the current throughresistor R2, after the switch has been closed for a very longtime?

___A

What is IC(0), the magnitude of the currentthrough the capacitor just after the switch is closed?

___A

An electron that has a horizontal initial speed of 4.50 106 m/senters a region that has an electric field of 120 V/m pointingvertically upward.
(a)How long does the electron take to travel 20 cm in thehorizontal direction?
____7.18E6_________s

(b) In this same time, what is the vertical displacement of theelectron? Treat up as positive.
________________cm

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I am having trouble the second portion of this problem. Heres mywork so far:

I calculated the force by F= (1.60E-19J)(120 V/m)=1.92E-17N

Then I calculated the acceleration: a= (1.92E-17N)/(9.11E-31kg)=2.1E13 m/s2 upward

Using the equation: Y=Voyt + (1/2)ayt^2

Y= (0) + .5(2.1E13 m/s^2)(4.44E-8)^2

Answer: 2.6 cm is wrong

PHYS 208 Final: PHYS 208 TAMU Final 2010 Solutions

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