Consider the following arrangement of potentials and a test charge (not pictured

Question

Consider the following arrangement of potentials and a test charge (not pictured)

located at infinity.

A. To which location will a test charge of +3

Consider the following arrangement of potentials and a test charge (not pictured) located at infinity. To which location will a test charge of +3 mu C travel? What will be its kinetic energy when it arrives? Find the capacitance of a capacitor that would store the same amount of energy as you found in part B. Suppose the +3 mu C test charge is replaced with a test charge of -3 mu C. To which location will it travel? If the new test charge has a mass of 27.3 x 10-21 kg, what will be its speed when it reaches its destination?

Explanation / Answer

PART A. A POSITIVE charge will move "downhill" toward the lowest potential (ie, the most negative voltage), which is - 7 V in this case.  

PART B. At an infinite distance, the potential V = 0.   The final potential is V = - 7 volts.    Both PE and KE at infinity are zero.

The change in energy in moving from infinity to V= -7 volts is    q (delta V) = (3x10^-6 C)(7 volts) = 2.1 x 10^ -5 Joule = final KE.

PART C.   The energy U stored in a capacitor is U = q^2 / (2C), or C = q^2 / (2 U ) = (3x10^-6C)^2 /(2x2.1x10-5J)= 2.14x10^ -7 Farads.

PART D. A NEGATIVE charge will move "uphill" to the highest potential, (ie, the most positive voltage), which is +4V in this case.

PART E. First, find KE, as in part B.     final KE = q (delta V) = (3x10^-6)(4V) = 1.2x10^-5 Joule

                 KE = (1/2) m v^2

     1.2x10^-5J = (1/2) (27.3x10^ -21 kg) v^2,.   solving for speed v, we get   v = 2.96x10^7 m/s.

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