Respuesta :
Answer:
Part a)
F = 14.4 N
Part b)
away from +6μC charge
Explanation:
Force on the charge placed midway between two charges is given by
[tex]F = \frac{kq_1q}{r^2} - \frac{kq_2q}{r^2}[/tex]
here we know that
[tex]q = 2\mu C[/tex]
[tex]q_1 = 6 \mu C[/tex]
[tex]q_2 = 4\mu C[/tex]
here the charge is placed at mid point so we have
[tex]r = 5 cm[/tex]
now we have
[tex]F = \frac{(9\times 10^9)(6\mu C)(2\mu C)}{0.05^2} - \frac{(9\times 10^9)(2\mu C)(4\mu C)}{0.05^2}[/tex]
[tex]F = 43.2 - 28.8 = 14.4 N[/tex]
Part b)
since +6μC charge is more in magnitude so the force due to this charge will be more so the net force on it is away from +6μC charge
A) The magnitude of the force on the test charge is; F = 14.384 N
B) The direction of the net force will be; away from the +6μC charge.
We are given;
q = +2μC = 2 × 10^(-6) C
q1 = +6μC = 6 × 10^(-6) C
q2 = +4μC = 4 × 10^(-6) C
r = 10/2 = 5 cm = 0.05 m (divided by 2 because the charge is placed halfway)
A) Magnitude of the force on the test charge is given by the formula;
F = (Kq1*q/r²) - (Kq2*q/r²)
Where k = 8.99 × 10^(9) N.m²/C²
Thus;
F = ((8.99 × 10^(9) × 6 × 10^(-6) × 2 × 10^(-6))/(0.05²)) - ((8.99 × 10^(9) × 4 × 10^(-6) × 2 × 10^(-6))/(0.05²))
F = 14.384 N
B) The charge of +6μC is greater than that of +4 μC in magnitude and therefore, the net force will be in a direction away from the +6μC charge.
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