The potential gradient between two charged plates having separation of 0.5cm and a potential difference of 12 volts is:
Correct answer: D. 2400 NC-1
- A. 240 NC-1
- B. 24 NC-1
- C. 2.4 NC-1
- D. 2400 NC-1
Explanation
As we know, Potential gradient is given by E= -dV/dR So, 12/0.005 = 2400 V/m= 2400 N/C
This question appeared on the ETEA MDCAT 2011 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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