Electric field strength at position vector r=(4i+3j)m caused by point charge q=5uC placed at origin is:
Correct answer: C. 1440i+1080jN/C
- A. 1440i+1080jV/m
- B. 1240i+1280jN/C
- C. 1440i+1080jN/C
- D. 1240i+1080jN/C
Explanation
The formula for the electric field strength is E=(K x Charge, Q)/distance2. The distance is the magnitude of the position vector which is d=(32 + 42)1/2. We multiply the electrostatic force by the unit vector which is the position vector, 4i + 3j, divided by the magnitude of the vector, 5 to get 1438.4i + 1078.8j. This rounds off to give C as the answer.
This question appeared on the UHS MDCAT 2017 paper, which you can sit online with every answer explained.
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About Electric Field Intensity
Electric field intensity is the force acting per unit positive test charge, with direction set by the force on that charge. Work includes fields of point charges, vector superposition, field lines and the relation between field and electric potential. Field intensity must be distinguished from electric force and potential.
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