example: calculate voltage inside, on the surface …srjcstaff.santarosa.edu/~yataiiya/4c/lecture...

18
I- on the surface: Lets choose points A and B on the surface Conclusion: Surface of any conductor is an equipotential surface Example: Calculate voltage inside, on the surface and outside a solid conducting sphere of charge Q PHYS42-9-3_10-2015-B Page 1

Upload: others

Post on 07-Apr-2020

6 views

Category:

Documents


0 download

TRANSCRIPT

I- on the surface: Lets choose points A and B on the surface

Conclusion: Surface of any conductor is an equipotential surface

Example: Calculate voltage inside, on the surface and outside a solid conducting sphere of charge Q

PHYS42-9-3_10-2015-B Page 1

II: inside the metallic sphere.

Conclusion: Entire body (not just the surface) of any conductor is an equipotential surface

III- outside metallic sphere:

Sphere acts as a point charge.

PHYS42-9-3_10-2015-B Page 2

Calculate and plot voltage everywhere for a non-conductive sphere of total charge Q and radius R

PHYS42-9-3_10-2015-B Page 3

Note: From previous examples we know that electric field inside a uniformly charged insulating sphere is

PHYS42-9-3_10-2015-B Page 4

See graphs generated by Excel below:

PHYS42-9-3_10-2015-B Page 5

Plot E and V everywherefor a point charge

Example 1:

Example 2-Spherical conductor

Let's look at Voltages again:

PHYS42-9-3_10-2015-B Page 6

Example 3- Conductive shell

Example 4: Nonconductive sphere

PHYS42-9-3_10-2015-B Page 7

Example 5: charge of Q inside a Conductive shell

PHYS42-9-3_10-2015-B Page 8

PHYS42-9-3_10-2015-B Page 9

Example: Voltage at a position in space is given as V=8X3Y+3Y2 +5ZX2. Calculate E (in component format) and magnitude of E at point 10,2,-3 (all in m)

Example: Part I. Charges q1 and q2 are separated by 30 cm. Calculate Electric field, voltage and force on Q=-8uc placed at point A.

PHYS42-9-3_10-2015-B Page 10

Part2: Two charges are brought together, touched and separated to the same distance. Repeat above

Let's assume L to R is

Positive

Q=-8uc placed at point A.

Since E is L to R and Q is negative:

F would be R to L

PHYS42-9-3_10-2015-B Page 11

Is it possible to have zero |E| but non-zero V and vice versa?

Part2: Two charges are brought together, touched and separated to the same distance. Repeat above calculations.

Since E is L to R, F is R to L (why?)

PHYS42-9-3_10-2015-B Page 12

Example: Two spherical conductors (R1=20 cm and R2=5 cm) have charges +35UC and -15 UC, respectively.

The spheres are connected together with a wire. Calculate charges and charge densities on each sphere.

At the middle of line, Point m

PHYS42-9-3_10-2015-B Page 13

We saw earlier that the voltage on the surface of a conductive sphere is Kq/R

Total charge on spheres 1 and 2 is (+35-15)=+20uC

V1 = V2

Concept Check: Q and are higher at sharper corner

PHYS42-9-3_10-2015-B Page 14

Combo Example!Two identical metallic spheres of small radiuses are separated from each other at a distance of 20 cm. Sphere to the left has a charge of -45UC and the sphere to the right has a charge of +15UC. A: Calculate Electric

field and voltage at the mid-point.

B-Calculate the force on a charge of q=-8UC placed at the mid-point

PHYS42-9-3_10-2015-B Page 15

C- Two spheres are brought together, momentarily touched, and separated to the same distance as before. Calculate electric field and voltage at the mid-point.

D-Calculate the force on a charge of q=-8UC placed at the mid-point

A positron (anti-electron) is injected into a capacitor (100V) thru a hole in its negative side. What should be the minimum speed of the particle to reach at least to the middle of the capacitor?

Voltage at the middle of the C is 200/2 = 100 V

Potential for the positron at this location is

Note: If plot of E versus distance is given, V can be calculated from negative of the area under the curve.

If plot of voltage versus distance is given, electric field can be calculated from slope of the graph.

PHYS42-9-3_10-2015-B Page 16

This is the end of topics being used for our first mid-term

Note: E due to charge distribution of infinite plane, conductive and insulators

Case 1: let's assume we place charge Q inside the infinite non-conductive plane. Assuming small thickness, we can define surface charge density,

=Q/A. As a result (using Gauss's law, see textbook) we can see that the |E| at the vicinity of

the plane would be /(20)

Case 2: This time, we place the same charge, Q on the surface of an infinite conductive plane. Each surface

PHYS42-9-3_10-2015-B Page 17

the plane would be /(20)

Case 2: This time, we place the same charge, Q on the surface of an infinite conductive plane. Each surface will have a total charge of Q/2, making the surface

charge density of each surface to be '=(Q/2)/A=/2.This time, the Gaussian surface will be different. The right surface of the cylinder will be inside the

conductor, so

=/(20)

Be careful when finding E of a thick solid conductor.

Since = Q/A, then E=/0

PHYS42-9-3_10-2015-B Page 18