Hi! could anyone help me please, I have no idea how to this at all. I try to it but, the did not get the right answer. Thank for your help
Oppositely charged parallel plates are separated by 5.34 mm. A potential difference of 600 V exists between the plates.
(a) What is the magnitude of the electric field between the plates?
___ N/C
(b) What is the magnitude of the force on an electron between the plates?
____ N
(c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.89 mm from the positive plate?
___ J
Did they mention the exact numbers of the charges of the parallel plates?
Field strength E times distance between plates = Voltage difference
Of course! I can help you with that.
To find the answers to these questions, we can use the formulas related to electric fields, electric forces, and work done.
(a) The formula for the magnitude of the electric field between two parallel plates is:
E = V/d
where E is the magnitude of the electric field, V is the potential difference between the plates, and d is the separation distance.
In this case, the potential difference is given as 600 V and the separation distance is given as 5.34 mm (which we can convert to meters by dividing by 1000).
Plugging these values into the formula, we get:
E = 600 V / (5.34 mm / 1000) = 112,359 N/C
So, the magnitude of the electric field between the plates is 112,359 N/C.
(b) The force on an electron in an electric field can be calculated using the formula:
F = qE
where F is the magnitude of the force, q is the charge of the particle, and E is the electric field.
The charge of an electron is q = -1.60 x 10^(-19) C (negative because it is an electron).
Plugging in the value of the electric field from part (a), we get:
F = (-1.60 x 10^(-19) C) * (112,359 N/C) = -1.80 x 10^(-14) N
So, the magnitude of the force on an electron between the plates is 1.80 x 10^(-14) N.
(c) The work done on a charged particle moving in an electric field is given by the formula:
W = qV
where W is the work done, q is the charge of the particle, and V is the potential difference.
Since we are moving an electron, the charge is q = -1.60 x 10^(-19) C (negative because it is an electron). And the potential difference is given as 600 V.
Plugging in these values, we have:
W = (-1.60 x 10^(-19) C) * (600 V) = -9.60 x 10^(-17) J
The negative sign indicates that work is done on the electron to move it to the negative plate.
So, the amount of work that must be done on the electron to move it to the negative plate is 9.60 x 10^(-17) J.
I hope this helps you! Let me know if there's anything else I can assist you with.