Maximum initial speed the electron can have without hitting the negative plate

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SUMMARY

The maximum initial speed (v0) of an electron launched at a 45-degree angle from a positive plate, without hitting a negative plate, is calculated to be 1.3 x 107 m/s. This conclusion is derived from the electric field strength of 1.0 x 104 N/C and the distance between the plates, which is 2.0 cm. The calculation involves determining the vertical component of the velocity (vy0) using the formula vy0 = sqrt(2 * E * q / m * ymax), where E is the electric field strength, q is the charge of the electron, and m is the mass of the electron.

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kirby2
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Two parallel plates are 2.0 cm apart and the electric field strength between
them is 1.0 × 10^4 N/C. An electron is launched at a 45 degree angle from the
positive plate. What is the maximum initial speed v0 the electron can have
without hitting the negative plate?

i think the answer is 1.3x10^7 m/s. can someone confirm this?
 
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kirby2 said:
Two parallel plates are 2.0 cm apart and the electric field strength between
them is 1.0 × 10^4 N/C. An electron is launched at a 45 degree angle from the
positive plate. What is the maximum initial speed v0 the electron can have
without hitting the negative plate?

i think the answer is 1.3x10^7 m/s. can someone confirm this?

It would be much easier for us to confirm it if you posted your work for us to see. Remember, that's part of the PF rules -- posting your attempt at the solution.
 
The acceleration = F/m = E*q/m

So at y max vy = 0 so

vy^2 = vy0^2 - 2*E*q/m*ymax

So vy0 = sqrt(2*E*q/m*ymax) = sqrt(2*1.20x10^4*1.60x10^-19/9.11x10^-31… = 9.18x10^6m/s

So v0 = vy0/sin(45) = 9.18x10^6m/s/sin(45) = 1.30x10^7m/s
 

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