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The relation between the electric field and the electrostatic potential is
A$E = +\dfrac{dV}{dr}$
B$E = -\dfrac{dr}{dV}$
C$E = V\,r$
D$E = -\dfrac{dV}{dr}$
Answer & Solution
Correct answer: D. $E = -\dfrac{dV}{dr}$
1. The field points in the direction of **decreasing** potential.
2. Quantitatively, $E = -\dfrac{dV}{dr}$ (the field is the negative gradient of potential).
3. The minus sign encodes that positive charges move from high to low potential.
_Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', Short Answer Q.13_
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