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The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between the charges is the same. The magnitude of the charges is |q|= 1.6 µC, and the distance between them is d = 2.6 mm. Determine the magnitude of the net force on charge 2 for each of the three drawings.

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okpalawalter8

Answer:

a)[tex]F = 6816.5680 N[/tex]

b) [tex]F' = 0 N[/tex]

c)[tex]F''=28195.5N[/tex]

Explanation:

Magnitude of charges [tex]M=1.6\muC[/tex]

Distance [tex]d=2.6[/tex]

Generally the equation for  Net Force is mathematically given by

For First Drawing

[tex]F =\frac{ k q^2}{d^2}+\frac{k q^2}{d^2}[/tex]  

[tex]F =\frac{2* 9*10^9* (1.6*10^-6)^2}{ (2.6*10^-3^2}[/tex]

[tex]F = 6816.5680 N[/tex]

For second Drawing

[tex]F' =\frac{ k q^2}{d^2 }-\frac{ k q^2}{ d^2}[/tex]  

[tex]F' = 0 N[/tex]

For Third Drawing

[tex]F'' =\frac{ k q^2}{d^2} * \sqrt (2)[/tex]

[tex]F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}[/tex]

[tex]F''=28195.5N[/tex]

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