Answer:
Part A: <em>4.74×10⁴ m</em>
P<em>art </em>B<em>: </em>0.000211 C
Explanation:
Part A:
From coulombs law,
F = kq²/r²........................ Equation 1
Note: The two object are carrying equal charges.
Where F = Electric force, q = charge, r = distance of separation, k = proportionality constant
<em>Given: F = 4.0 N, q = 1.0 C, </em>
<em>Constant: k = 1/4πe₀ = 9×10⁹ Nm²/C²</em>
<em>Substituting these values into equation 1,</em>
<em>4 = 9×10⁹(1²)/r²</em>
<em>r² = (9×10⁹)/4</em>
<em>r² = 2.25×10⁹</em>
<em>r = √(2.25×10⁹)</em>
<em>r = 4.74×10⁴ m.</em>
<em>Part 2:</em>
<em>F = kq²/r²</em>
<em>Making q the subject of the equation</em>
<em>q = √(Fr²/k)........................... Equation 2</em>
Where: F = 4.0 N, r = 1.0 m, k = 9×10⁹ Nm²/c²
Substituting these values into equation 2
q = √(4×1²/(9×10⁹ )
q = 2.11/10⁴
q = 0.000211 C.
