# Electric field strength and potential gradient

by zinc79
Tags: electric, field, gradient, potential, strength
 P: 9 A bit of a problem. My book teaches me that E = -(dV/dx), where E is the electric field strength, V is the electric potential, and x represents displacement. But, it also suggests along with the above formula that E = -(V/d) and displays a circuit with a battery of p.d. V and two parallel metal plates of distance (d) from each other. My question is, HOW did E = -(dV/dx) become E = -(V/d)??? The former formula, proven via differentiation, says that the electric field strength is negative of the potential gradient i.e. rate of change of electric potential with respect to the displacement. Then how does this transform into the electric field strength simply being equal to the negative of the ratio of the potential difference to the distance? It makes no sense to me! And yet, I've seen the latter formula being used in practice questions.
 P: 1,133 If the rate of change of V is constant over the distance x (or d), then -dV/dx can be stated as -V/d. -V/d is not equal to -dV/dx, but in a parallel plate capacitor it usually is.
 Sci Advisor HW Helper PF Gold P: 4,108 To follow up on Gear300's comment... the general relation is E = -(dV/dx)... for parallel plates, one really should write Eparallel plates = -(V/d) Recall for parallel plates, Eparallel plates is a constant [between the plates]. (Don't merely match symbols or abbreviations. Not all "E"s mean the same thing. For each "formula", understand what the variable represents.)
P: 9

## Electric field strength and potential gradient

 Mentor P: 11,231 The parallel-plate equation $E = -V/d$ really should read $E = -\Delta V/d$ to indicate that the voltage is the difference in potential between the two plates. Unfortunately it's a long-standing custom to use V to represent both potential (at a point) and potential difference (between two points) in different contexts.
 Quote by jtbell The parallel-plate equation $E = -V/d$ really should read $E = -\Delta V/d$ to indicate that the voltage is the difference in potential between the two plates. Unfortunately it's a long-standing custom to use V to represent both potential (at a point) and potential difference (between two points) in different contexts.