A capacitor with plates separated by 0.0290 m is charged to a potential difference of 5.50 V. All wires and batteries are then disconnected, and the two plates are pulled apart to a new separation distance of 0.0580 m. If the initial electric field strength is 190. N/C, what is the electric field after the increase in separation? 95.0 N/C 190. N/C 760. N/C 380. N/C
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Step 1: The electric field strength is given by the formula E = Q / (ε0 * A), where Q is the charge on the plates, ε0 is the permittivity of free space, and A is the area of the plates. Show more…
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