What is the total charge of the plutonium nucleus? (The neutral plutonium atom h

Question

What is the total charge of the plutonium nucleus? (The neutral plutonium atom has 94 electrons.)
1.50×10-17 C

a)What is the magnitude of the force on an electron at a distance of 7.46 angstrom from the plutonium nucleus?

b)What would the magnitude of the force be if the distance of the electron from the nucleus were tripled?

2. Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.290 m to the right of Q1. Q3 is located 0.136 m to the right of Q2. The force on Q2 due to its interaction with Q3 is directed to the.....
True: Left if the two charges are negative.
False: Left if the two charges have opposite signs.
True: Left if the two charges are positive.
True: Right if the two charges have opposite signs.
False: Right if the two charges are negative.

a)In the above problem, Q1= 2.07·10-6 C, Q2= -2.65·10-6 C, and Q3= 3.03·10-6 C. Calculate the total force on Q2. Give with the plus sign for a force directed to the right.

b)Now the charges Q1= 2.07·10-6 C and Q2= -2.65·10-6 C are fixed at their positions, distance 0.290 m apart, and the charge Q3= 3.03·10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.

Explanation / Answer

1)

a)charge q1=1.5*10^-17c

Q2=1.6*10^-19c

Distance between the charges=7.46 x10^-10m

The force repulsion between themis

F=(1/4pie0)(1.5*1.6*10^-19*10^-17)(1/(7.46*10^-10)^2)

=3.88*10^-8N

B)if the new distance =3*7.46*10^-10m

Then force of repulsion between them is equal to(1/9) of the previous force

F"=(1/9)(3/88*10^-8)

=4.31*10^-9N

2)a)

Given Q1=2.07*10^-6c

Q2=-2.65*1.0^-6c

Q3=3.03*20^-6c

Distance between q1&,q2 pis 0.29 m

Distance between q3&,q2 pis 0.136m

Net force on q2 is=(1/4pie0)((2.07*2.65)/0.39^2)10^-12-(1/4pe0)((3.03*2.65)/0.136^2)10^-12

On solving we get

Net force on q2 is -3.32N acting towards left

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