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A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

User Zxcv
by
4.4k points

1 Answer

1 vote

Answer:

F = 37.8 × 10^(6) N

Step-by-step explanation:

The charges are 0.06 C and 0.07 C.

Thus;

Charge 1; q1 = 0.06 C

Charge 2; q2 = 0.07 C

Distance between them; r = 3 m

Formula for the force in between them is;

F = kq1•q2/r²

Where k is a constant = 9 × 10^(9) N.m²/C²

Thus;

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

User Harwalan
by
4.7k points