Speed of an Electron in an Electric Field

AI Thread Summary
The discussion revolves around calculating the final speed of an electron released from rest between two positive point charges. The problem involves using conservation of energy principles, where the initial potential energy is converted into kinetic energy as the electron moves. Relevant equations include those for electric potential from point charges and the relationship between potential energy and kinetic energy. The user is guided to sum the electric potentials from both charges and apply energy conservation to find the final speed of the electron. The key takeaway is to carefully manage the signs and values in the equations to arrive at the correct final speed.
HELLO11
Messages
19
Reaction score
0

Homework Statement



Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 1.55nC , are separated by a distance of 57.0cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

find Vfinal of the electron when it is 10.0 cm from charge 1 the want the final answer in m/s

i don't get any of this my professor went to fast. can anyone help?
 
Physics news on Phys.org
This looks like a conservation-of-energy problem.

The two relevant forms of energy are kinetic and electric potential.
 
do you know of any equations that could help me
 
HELLO11 said:
do you know of any equations that could help me

Your book or class notes will have an equation for the electric potential (or voltage) due to a point charge.

To get the net electric potential of the two stationary charges, add up the individual potentials of each of the two charges.

To get the potential energy, multiply electric potential times the charge on the electron. Watch the +/- signs carefully.
 
can anyone help me
 
q1 = 3.25 nC, q2 = 1.55nC, d = 0.570 m. charge of electron = q, a = 0.100 m, find the final speed of the electron, v = ?
initial kinetic energy Eki = 0, potential energy Epi = Kqq1/(d/2) + Kqq2/(d/2)
final kinetic energy Ekf = mv^2/2, potential energy Epf = Kqq1/a + Kqq2/(d-a)
energy conservation:
0 + Kqq1/(d/2) + Kqq2/(d/2) = mv^2/2 + Kqq1/a + Kqq2/(d-a)
solve for v.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top