The electric field at a point due to two equal and opposite charges is 100 N/C. If the magnitude of each charge is doubled then the electric field at that point becomes:
Correct answer: D. 400 N/C
- A. 50 N/C
- B. 100 N/C
- C. 200 N/C
- D. 400 N/C
Explanation
This is the option marked correct in the official UHS answer key. Field intensity is directly proportional to the charge producing it, so doubling both charges doubles each contribution and doubles the resultant, which by that reasoning gives 200 N/C. The key's figure of 400 corresponds to treating the effect as multiplying by four, so this is a question where the official answer and the standard proportional argument diverge.
This question appeared on the UHS MDCAT 2025 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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