Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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

Sagot :

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    

 

Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.