ch 25.5 – potential due to continuous distribution we have an expression for the e-potential of a...

15
Ch 25.5 – Potential due to Ch 25.5 – Potential due to Continuous Distribution Continuous Distribution We have an expression for the E-potential of a point charge. Now, we want to find the E-potential due to weird distributions of charge. r q k V e Electric potential due to a point charge.

Upload: olivia-banks

Post on 18-Dec-2015

216 views

Category:

Documents


2 download

TRANSCRIPT

Ch 25.5 – Potential due to Continuous DistributionCh 25.5 – Potential due to Continuous Distribution

• We have an expression for the E-potential of a point charge.

• Now, we want to find the E-potential due to weird distributions of charge.

r

qkV e

Electric potential due to a point charge.

Ch 25.5 – Potential due to Continuous DistributionCh 25.5 – Potential due to Continuous Distribution

• Approach is similar to that for calculating E-fields.

• Break down the overall distribution into little chunks, find the E-potential due to each, and add up the contributions to get

r

dqkV e

Electric potential due to a continuous charge distribution.

r

dqkdV e

(a) Find an expression for the electric potential at a point P located on the central axis of a uniformly charged ring of radius a and total charge Q.

(b) Find an expression for the magnitude of the E-field at point P.

EG 25.5 The electric potential due to a ringEG 25.5 The electric potential due to a ring

A uniformly charged disk has radius R and surface charge density σ.

(a) Find the electric potential at a point P located on the disk’s central axis.

(b) Find the x component of the E-field at P along the central axis of the disk.

EG 25.6 The electric potential due to a diskEG 25.6 The electric potential due to a disk

A rod of length l located on the x axis has a total charge of Q and a uniform linear charge density of λ = Q/l. Find the E-potential at point P located on the y axis a distance a above the origin.

EG 25.7 The electric potential due to a lineEG 25.7 The electric potential due to a line

Hint: 22

22ln axx

ax

dx

Already know:

1) If a conductor carries net charge, it resides on the surface.

2) The E-field just outside the conductor is perpendicular to the surface.

Now, we show:

Every point on the surface of a charged conductor in equilibrium is at the same electric potential.

Ch 25.6 – E-Potential Due to a Charged ConductorCh 25.6 – E-Potential Due to a Charged Conductor

Consider two points, A and B, on the surface of a charged conductor.

Ch 25.6 – E-Potential Due to a Charged ConductorCh 25.6 – E-Potential Due to a Charged Conductor

0 B

A

AB sdEVVV

Since the E-field is always perpendicular to the surface, any path from A to B along the surface will generate zero potential difference:

Ch 25.6 – E-Potential Due to a Charged ConductorCh 25.6 – E-Potential Due to a Charged Conductor

The surface of any charged conductor in equilibrium is an equipotential surface.

Because the E-field is zero inside the conductor, the electric potential is constant everywhere inside the conductor and equal to its value at the surface.

Ch 25.6 – E-Potential Due to a Charged ConductorCh 25.6 – E-Potential Due to a Charged Conductor

The electric potential and the electric field due to a positively charged, spherical conductor.

1) Charge is smeared evenly over the surface.

2) Electric potential is constant inside the sphere and acts like the potential due to a point charge outside the sphere.

3) The E-field is zero inside the sphere, and acts like a positive point charge outside the sphere.

Ch 25.6 – E-Potential Due to a Charged ConductorCh 25.6 – E-Potential Due to a Charged Conductor

Recall, surface charge density on a conductor is determined by the radius of curvature.

Small radius of curvature (very pointy) high surface charge density.

Large radius of curvature (flat) low surface charge density.

EG 25.8 Two connected charged spheresEG 25.8 Two connected charged spheres

Two spherical conductors of radii r1 and r2 are separated by a distance much greater than the radius of either sphere. The spheres are connected by a conducting wire. The charges on the spheres in equilibrium are q1 and q2, respectively.

Find the ratio of the magnitudes of the E-fields at the surfaces of the spheres.

Ch 25.6 – Cavity within a conductorCh 25.6 – Cavity within a conductor

Suppose a conductor contains a cavity.

Assume the cavity contains no charges.

The E-field inside the cavity must be zero, even if an E-field exists outside the conductor.

Ch 25.6 – Cavity within a conductorCh 25.6 – Cavity within a conductor

Remember, every point on the conductor is at the same potential.

The only way for the potential difference to be zero from A to B is if the E-field at all points inside the cavity is zero.

0 B

A

AB sdEVVV

Must be zero.

Ch 25.6 – Cavity within a conductorCh 25.6 – Cavity within a conductor

A cavity surrounded by conducting walls is a field-free zone, as long as no charges are inside the cavity.

Ch 25.6 – Corona dischargeCh 25.6 – Corona discharge

Corona discharge often occurs near the surface of a charged conductor, like a high voltage power line.

A high potential difference between the conductor and the surrounding air can strip electrons from the air molecules.

The electrons accelerate, colliding with other air molecules. The result is the emission of light.

Corona Discharge