The electric field between two parallel plates is uniform, with magnitude 646 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.26 cm. At the same moment, both particles are released. Required:
Determine the distance from the positive plate at which the two pass each other.

The distance from the positive plate at which the two pass each other is 0.0023 cm.

Explanation:

We need to find the acceleration of each particle first. Let's use the electric force equation.

For the proton

For the electron

Now we know that the plate separation is 4.26 cm or 0.0426 m. The travel distance of the proton plus the travel distance of the electron is 0.0426 m.

Both of them have an initial speed equal to zero. So we have:

With this time we can find the distance from the positive plate (x(p)).

Therefore, the distance from the positive plate at which the two pass each other is 0.0023 cm.

I hope it helps you!

i think the correct answer from the choices listed above is option c. when the kinetic energy of particles in a liquid state decreases

explanation:

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz