With two points i and f separated by a small distance, where the distance is essential constant, what is the work done by the electric field as charge q moves through the small distance

W= Fs x deltas= qEs x delta s

Potential difference between those two-point charges at i and f

delta V= Uq+othersources/q= -W/q= -(qEsdeltas)/q= -Es x deltas

In terms of potential, the component of the electric field in the s-direction is and how is the field in relation to the coor axis

Es= deltaV/deltaS as the limit is s->0 meaning Es= -dV/dS parallel

Other relation of determining the component of the electric field in the s-direction and how to deal with it

1) Use symmetry to select a coor axis in parallel to Efield and along which the perpendicular component of E is 0

Knowing potential of a point charge to get electric field with symmetry and radical component

Symmetry requires electric field to point directly outward from point charge so radical component is Er and if s-axis is to be in this radical direction, directly outward as in parallel to Efield, we get V=q/4pieeor Er= -dV/ds= -d/dr (1=q/4pieeor)= 1/4pieeo x q/r^2 (electrical field of point charge)

Usefulness of the efield equation

Es= -deltaV/deltas as s->0 as the limit then Er= -dV/ds Useful for a continuous distribution of charge as calculating V is easier than the vector Efield directly from charge as it's a scalar son once its known, then the derivative is the efield

Another way of looking at the field equation

Es= -dV/ds Electric field is the negative slope of a V-versus-s graph

Force on a particle is what of the slope of a potential- energy graph and relation with efield equation

Negative so F= -dU/ds Relation is just dividing both sides by q to get just E and V

Potential in terms of Es and Es in terms of Potiential importance

deltaV= Vf- Vi= integral of si to sf (Esds)= integral of i to f (vector E x vector ds) & Es= delta V/ delta s Importance in terms of geometry for potential and field

Electric potential at P as related to surfaces of equipotiential surfaces

With two differently charged equipotential surfaces, allow point charge to move into two displacements deltas1 vector & deltas2 vector

Efield of first displacement vector on two equipotential surfaces

The first displacement is tangent to equipotential surface V-, so it experiences no potential difference, so it is in a direction so constant potential as it now has the electric field vector that's tangent to the equipotential surface is 0

Efield of second displacement vector on two equipotential surfaces

Perpendicular to equipotential surface in the direction of increasing potential (V- ->V+) Potential difference exists so electric component is Eperpendicular= -dV/ds= deltaV/delta s= -(V+ - V-)/delta s2

Conclusions for Efield from the Efield of second displacement vector on two equipotential surfaces

- Electric field is inversely proportional to delta s, distance between equipotential surfaces - As (V+ - V-)> 0, the minus in the original Efield equation tells us direction of Efield is opposite to direction of delta s meaning it is perpendicular everywhere to the surfaces in the downhill direction of decreasing potiential, V --Equipotential surfaces have equal potential differences between them

Individual Components of E at any point by extending efield equation

E= -delta v/ delta s= Exi +Exj + Exk= ((partial derivative of V in respect to x)i + (partial derivative of V in respect to x) j + (partial derivative of V in respect to) k x, y, z held constant where the partial derivatives in () are the gradient of V so Efield= -delta V (gradient of V)

With two points, 1 and 2, in a region of both potential and a field, the work done in moving a charge between points 1 and 2 is what of the path

Independent

For any path that ends and starts at the same point regardless of the differences in potentials and the number of charges all in a closed circuit

Net Potential will always be 0V as deltaVloop= dum of i (deltaV)i= 0 So, the sum of all potential differences encountered while moving around a loop or a closed path is 0