Electric field strength and speed

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The electric field strength in the parallel-plate capacitor is 1.70×10^4 N/C with a spacing of 1.40 m. An electron released from rest at the negative plate will accelerate towards the positive plate due to the electric field. To find the electron's speed upon reaching the positive plate, one can apply Newton's second law and kinematic equations. Specifically, the force acting on the electron can be calculated using F = ma, and the final speed can be determined using the equation (v_f)^2 = (v_i)^2 + 2ad. Combining these principles allows for the calculation of the electron's speed effectively.
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The electric field strength is 1.70×104 inside a parallel-plate capacitor with a 1.40 spacing. An electron is released from rest at the negative plate.

What is the electron's speed when it reaches the positive plate?
Help please...
 
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You need to combine Forces with Kinematics. We know that F = KE/r and also know that (v_{f})^2 = (v_{i})^2 + 2ad and of course, Newton's second law: F = ma. You have everything you need, just use some basic algebra, and that's it.
 
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