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

Transcription

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

2 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_ B Page 2

3 PHYS42-9-3_ B Page 3 Calculate and plot voltage everywhere for a nonconductive sphere of total charge Q and radius R

4 Note: From previous examples we know that electric field inside a uniformly charged insulating sphere is PHYS42-9-3_ B Page 4

5 See graphs generated by Excel below: PHYS42-9-3_ B Page 5

6 Let's look at Voltages again: Example 1: Plot E and V everywhere for a point charge Example 2-Spherical conductor PHYS42-9-3_ B Page 6

7 Example 3- Conductive shell Example 4: Nonconductive sphere PHYS42-9-3_ B Page 7

8 Example 5: charge of Q inside a Conductive shell PHYS42-9-3_ B Page 8

9 PHYS42-9-3_ B Page 9

10 Example: Voltage at a position in space is given as V=8X 3 Y+3Y 2 +5ZX 2. 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_ B Page 10

11 Q=-8uc placed at point A. Let's assume L to R is Positive Since E is L to R and Q is negative: F would be R to L Part2: Two charges are brought together, touched and separated to the same distance. Repeat above PHYS42-9-3_ B Page 11

12 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?) Is it possible to have zero E but non-zero V and vice versa? PHYS42-9-3_ B Page 12

13 At the middle of line, Point m 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. PHYS42-9-3_ B Page 13

14 Total charge on spheres 1 and 2 is (+35-15)=+20uC V1 = V2 We saw earlier that the voltage on the surface of a conductive sphere is Kq/R Concept Check: Q and are higher at sharper corner PHYS42-9-3_ B Page 14

15 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_ B Page 15

16 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_ B Page 16

17 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 /(2 0 ) Case 2: This time, we place the same charge, Q on the surface of an infinite conductive plane. Each surface PHYS42-9-3_ B Page 17

18 the plane would be /(2 0 ) 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 = /(2 0 ) Be careful when finding E of a thick solid conductor. Since = Q/A, then E= / 0 PHYS42-9-3_ B Page 18