Chapter 16 Potential Energy Potentials Equipotential Sfcs Remember

Chapter 16 Potential Energy Potentials Equipotential Sfcs.

Remember, the change in potential energy is minus the work done by the field force. What happens to the potential of a charge as it moves in the direction of the electric force? +q E d Va Vb

+q E d Va Vb Vb - Va = - E d It decreases in the direction of the electric field, REGARDLESS OF THE SIGN OF THE CHARGE!

+q A Fe d B E What happens to the potential energy of a positive charge (+q) as it moves in the direction of the electric field? DPE = - q E d = - (+q) E d = - q E d

-q A Fe d B E What happens to the potential energy of a negative charge (-q) as it moves in the direction of the electric field? DPE = - q E d = - (-q) E d = + q E d

+q A Fe B d E What happens to the potential of a positive charge (+q) as it moves in the direction of the electric field? DV = - E d

-q A Fe B d E What happens to the potential of a negative charge (-q) as it moves in the direction of the electric field? DV = - E d

So, how’s this work for point charges? First, assume that the potential in the field of a point charge is ZERO at infinity (we did something similar for gravity, right? ) Use Calculus…. VOILA! scalar!

V>0 around positive charges V<0 around negative charges The superposition principle applies to potentials! Vtot = V 1 + V 2 + V 3 +. . .

How much work do we do in bringing a point charge (q 2) from infinity to a distance r from point charge q 1? q 1 r q 2

PE > 0 for LIKE charges PE < 0 for opposite charges You could probably have guessed that PEtot = PE 1 + PE 2 + PE 3 +. . .

The electrical potential ALWAYS decreases in the direction of the electric field! It does not depend upon the sign of the charge. The electrical potential energy depends upon the sign of the charge. It decreases in the direction of the electrical force.

What is the potential at point p? p 1 m q 1 Vtot = V 1 + V 2 2 m q 1 = +2 m. C q 2 = -5 m. C q 2

p What is the potential at point p? 1 m q 1 2 m q 1 = +2 m. C q 2 = -5 m. C q 2 Vtot = V 1 + V 2 V 1 = 18, 000 V V 2 = -20, 125 V Vtot = -2, 125 V

p 1 m q 1 2 m q 1 = +2 m. C q 2 = -5 m. C q 2 How much work is done by the electric field to bring a 5 m. C charge from infinity to point p? W= -(5 X 10 -6 C) (-2, 125 V) = +1. 1 X 10 -2 J

An equipotential surface is a set of points in an electric field which are all at the same electrical potential. One example is the locus of points equidistant from an isolated point charge. Electric field lines Equipotential surfaces

Notice that the equipotential surfaces are perpendicular to the electric field lines everywhere! Remember that work is only done when a charge moves parallel to the electric field lines. So no work is done by the electric field as a charge moves along an equipotential surface. This is just like GRAVITY, right?

The potential difference between any two points on an equipotential surface (Vb - Va ) must be. . .

When in electrostatic equilibrium (i. e. , no charges are moving around), all points on and inside of a conductor are at the same electrical potential!

Remember that the electric field is everywhere perpendicular to the surface of a conductor, and is zero inside of a conductor. So, from the surface of a conductor throughout its interior the electric field is 0 when the conductor is in electrostatic equilibrium. DV = - E d = 0