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A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an electric field. What are the magnitude and direction of the electric field at the point in question

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okpalawalter8

Answer:

[tex]E = -6 \ N/C[/tex]

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

Explanation:

From the question we are told that

   The charge on the small object is [tex]Q = -4.00 \ nC = -4.00 *10^{-9} \ C[/tex]

   The force is  [tex]F = 24 \ nN = 24 *10^{-9} \ N[/tex]

    Generally the magnitude of the electric  field is mathematically represented as

       [tex]E = \frac{F}{Q}[/tex]

=>    [tex]E = \frac{ 24 *10^{-9}} {-4 *10^{-9 }}[/tex]

=>     [tex]E = -6 \ N/C[/tex]

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

 

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