If the magnitude of the electric field at a distance of 7.8 cm from the center is 30500 n/c , what is the magnitude of the electric field at 22.3 cm from the center? answer in units of n/c.

3730 N/C The intensity of the electric field follows the inverse square law. Since everything except the distance is remaining constant, then the force will vary with the inverse square of the ratio of the change. So first, lets calculate the ratio of the change in distance. 22.3 / 7.8 = 2.858974359 And since we actually want the square of the distance... 2.858974359^2 = 8.173734385 And since the distance increased, that means the force will decrease. So we get 30500 / 8.173734385 = 3731.464538 Rounding to 3 significant figures gives 3730 N/C.

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