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An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What are the magnitude and direction of the electric field?

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MissPhiladelphia
The direction of the electric field would be south. 

qE/m = 115 
       E = 115*m/q 
           = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) 
           = 762.87 * 10^(-12) 
           = 6.27 x 10^-10 N/C

Hope this answers the question. Have a nice day. Feel free to ask more questions.

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