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Writer's picturekeshprad

PhysicsBowl 2016 Q25


25) In this question, we need to find the magnitude of the charge released in the situation described. We have a charge that experiences a 1.0 Newton electric force after being released between the plates of a capacitor.

Where V is potential difference, F is electric force, q is electric charge, and d is capacitor plate separation

From this point, we can come up with an equation for the electric force. You will see that we get an equation for force that includes E(electric field). We need to convert this to values that are known. Then, we are able to solve for the charge.

Where F is electric force, q is electric charge, E is electric field, V is potential difference, and d is capacitor plate separation

Answer: A

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