A proton is placed in an electric field of intensity 500 N/C. What are the magnitude...
| 4.79 � 109 m/s2 opposite to the electric field | |
| 47.9 � 1010 m/s2 opposite to the electric field | |
| 47.9 � 1010 m/s2 in the direction of the electric field | |
| 4.79 � 1010 m/s2 opposite to the electric field | |
| 4.79 � 1010 m/s2 in the direction of the electric field |
Homework Answers
1. a = Eq/m = 500 * 1.6*10^-19/ 1.67 *10^-27 = 4.79 *10^ 10 m/s^2
in the direction of electric field
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