Solving for a with Ma=Eq, ΔV, and m/d

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To solve for the acceleration of a charge q between two parallel plates with a potential difference ΔV, the electric field E can be related to ΔV and the plate separation d using the equation E = ΔV/d. The approach of using ma = Eq is correct, but the missing link is the relationship between the electric field and the potential difference. By substituting E with ΔV/d, the equation can be rearranged to find the acceleration a = qΔV/md. Understanding this relationship is crucial for deriving the correct expression for acceleration.
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Homework Statement
Two parallel plates are separated by a distance d and a potential difference ΔV is applied between them. Write an expression for the acceleration of a charge q with a mass m between the plates
Relevant Equations
E=F/q, C=q/V, E=1/2CV^2
Tried using ma=Eq, then subbing E for 1/2CV^2, but can't seem to be getting anywhere. How can I get to the answer which is a=qΔV/md.
 
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alanwangcot said:
Homework Statement: Two parallel plates are separated by a distance d and a potential difference ΔV is applied between them. Write an expression for the acceleration of a charge q with a mass m between the plates
Homework Equations: E=F/q, C=q/V, E=1/2CV^2

Tried using ma=Eq, then subbing E for 1/2CV^2, but can't seem to be getting anywhere. How can I get to the answer which is a=qΔV/md.
You could find the electric field.
 
ma=Eq is the right approach. The reason you aren’t getting anywhere is because you are missing an equation. If they told you the capacitance you’d be in business, but they didn’t. They gave you the plate separation d. Can you relate the Electric field to the potential and the plate separation? (hint: it’s kinda definitional)
 
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