ELECTRIC POTENTIAL. September 13, 2006. Goings On For the Next Few Days. Quiz Today – Gauss/Electric Field Today – Begin Chapter 25 – Potential Monday - Potential Wednesday – EXAMINATION #1 Friday – More Potential. Quizzicle. Picture a Region of space Where there is an Electric Field.
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September 13, 2006
Push vs Pull
Mrs. FIELDS vs Mr. External
The net work done by a conservative (field)
force on a particle moving
around a closed path is
Work done by external force = mgh
How much work here by
HIGH U LOWER U
Work done by FIELD is Fd
Negative of the work done by the FIELD is -Fd
Change in Potential Energy is –Fd.
The charge sort-of “fell” to lower potential energy.
We define ELECTRICAL POTENTIAL as the POTENTIAL ENERGY PER UNIT CHARGE:
If we move a particle through a potential difference of DV, the work from an external “person” necessary to do this is qDV
The difference in potential between the accelerating plates in the electron gun of a TV picture tube is about 25 000 V. If the distance between these plates is 1.50 cm, what is the magnitude of the uniform electric field in this region?
An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 × 10–17 J. Calculate the charge on the ion.
It takes NO work to move a charged particle
between two points at the same potential.
The locus of all possible points that require NO
WORK to move the charge to is actually a surface.
Work to move a charge a distance
Dx along the equipotential surface
Is Q x Eparallel X Dx
VConsider Two EquipotentialSurfaces – Close together
Work to move a charge q from a to b:
Note the (-) sign!
Over a certain region of space, the electric potential is V = 5x – 3x2y +2yz2. Find the expressions for the x, y, and z components of the electric field over this region. What is the magnitude of the field at the point P that has coordinates (1, 0, –2) m?
1 N/C = 1 V/m
Consider a unit charge (+) being brought from infinity to a distance r from a
To move a unit test charge from infinity to the point at a distance r from the charge q, the external force must do an amount of work that we now can calculate.
Set the REFERENCE LEVEL OF POTENTIAL
at INFINITY so (1/rA)=0.
Given two 2.00-μC charges, as shown in Figure P25.16, and a positive test charge q = 1.28 × 10–18 C at the origin, (a) what is the net force exerted by the two 2.00-μC charges on the test charge q? (b) What is the electric field at the origin due to the two 2.00-μC charges? (c) What is the electrical potential at the origin due to the two 2.00-μC charges?
The three charges in Figure P25.19 are at the vertices of an isosceles triangle. Calculate the electric potential at the midpoint of the base, taking q = 7.00 μC.
A disk of radius R has a non-uniform surface charge density σ = Cr, where C is a constant and r is measured from the center of the disk. Find (by direct integration) the potential at P.
q1=12nC q2=-24nC q3=31nC q4=17nC Sq=36 x 10-9C
V=350 Volts (check the arithmetic!!)
In the figure, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?
Derive an expression in terms of q2/a for the work required to set up the four-charge configuration in the figure, assuming the charges are initially infinitely far apart.