Edexcel Physics Unit 4.4 Electric Fields

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  • Created by: Reductito
  • Created on: 05-09-13 12:57

Proving that NC^-1 = Vm^-1

Electric Field Strength (E) is measured in Newtons per coulomb (NC^-1). However it can also be calculated as the gradient on a graph of potential difference over total distance travelled from plate (Vm^-1). [Context - this equation relates to an experiment where two metal plates are attached to the sides of a material, one plate is positively charged, one negatively charged. In the experiment the potential difference of the material changes depending how far away from the positve plate you get.]

So E = NC^-1. E also = Vm^-1. But why?

For this next part NC^-1 shall be represented as N / C to prevent complications.

Dimensional analysis

E = N / C.

N is a Newton which is the measurement for force (F).

C is a Coulomb which is the measurement for charge (Q).

E = F / Q.

Q = It. I is measured in amps (A) and t is measured in seconds (s). So It = As. Q = As.

F = W/d (derived from the equation W = Fd). W is measured in joules (J) and d is measured in metres (m). So W/d = Jm^-1. F = Jm^-1.

Now we input our units for force (F) and charge (Q) into F / Q.

E = Jm^-1 / As.

J = a joule. A joule is the unit for energy. A formula for calculating energy is E = VIt.

So J = VIt.

The units of potential difference (V) are volts (V). The units of current (I) are amps (A). The units of time (t) are seconds (s).

So J = VAs.

Putting this back into the equation E = Jm^-1 / As.

E = VAsm^-1 / As.

The As on each part of the equation cancels out.

E = Vm^-1.

Therfore, E = Vm^-1 which also = NC^-1.

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