physics

In the previous problem (Question 4), if the distance between the two plates of the capacitor is 3 cm, what is the magnitude of the uniform electric field between the two plates?

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  1. d = .03 meters
    E = V/d

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  2. so... E=.03J/3m=0.01J/m??

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  3. no
    I do not know what question 4 is
    but divide the voltage by .03 meters to get E

    in a capacitor E = volts/ distance

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  4. The electric potential increases from 10 V to 70 V from the bottom plate to the top plate of a parallel-plate capacitor. We are going to move a charge of +5 x 10-4 C from the bottom plate to the top plate. What is the magnitude of the change in potential energy of this charge? Do not enter any (-) sign in your answer.

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  5. so E=v/d

    E=.03/3
    E=.01?

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  6. V = 60 volts
    d = .03 meters

    E = 60/.03 = 2000

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