A small glass bead has been charged to 8.0 nc. what is the magnitude of the electric field 2.0 cm from the center of the bead?

1 answer:

7 0

<span>Charge of the glass bead Q = 8.0 x 10^-9 C Distance d = 2.0 cm = 0.02 m Coulombs constant K = 8.99 x 10^9 Nm^2/C^2 Electric Field E = k x Q / d^2 = 8.99 x 10^9 x 8.0 x 10^-9 / (0.02)^2 E = 71.92 / 0.0004 = 17.98 x 10^4 The electric field is 1.8 x 10^5 N/C</span>

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