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A test charge of +2 μC is placed halfway between a charge of +6 μC and another of +4 μC separated by 10 cm. What is the magnitude of the force on the test charge?

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MissPhiladelphia
For the answer to the question above, It is the difference between 6 mu C and 4 mu C. Both are repulsive. Therefore, the resultant force will be away from 6 muC to 4 muC. Because of this 2 muC will be pushed towards 4muC.
i hope this helps
asifejaz20

Answer:

Force on the test charge = F = 14.4N

Explanation:

1st charge = q1 = +6uc  

2nd charge = q2 = +4uc

Test charge = q0 = +2uc

Coulomb’s constant = [tex]k = 9*10^9 Nm^2/C^2[/tex] k

Distance between q1 and q2 = r12 = 10cm = 0.1m  

Distance of test charge from each other charge = r = 5cm = 0.05m

Distance of test charge from each other charge = r = 5cm = 0.05m  

As all three charges are positive so, there will be force of repulsion between them. Test charge is placed between q1 and q2 so, the total force will be subtracted. So,  

 [tex]F = F1 - F2[/tex]

[tex]F = kq1q0/r^2 - kq1q0/r^2[/tex]   [tex]F = (9*10^9)(6u)(2u)/(0.05)^2 - (9*10^9)(4u)(2u)/(0.05)^2[/tex]  

[tex]F = 14.4N[/tex]  

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