# Chapter 18.2 Review Capacitance and Potential - PowerPoint PPT Presentation

PPT – Chapter 18.2 Review Capacitance and Potential PowerPoint presentation | free to download - id: 748d80-MjA3Z

The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
Title:

## Chapter 18.2 Review Capacitance and Potential

Description:

### Title: It is sometimes difficult to find the polarity of an induced emf. The net magnetic field penetrating a coil of wire results from two factors. – PowerPoint PPT presentation

Number of Views:28
Avg rating:3.0/5.0
Slides: 36
Provided by: KentH164
Category:
Tags:
Transcript and Presenter's Notes

Title: Chapter 18.2 Review Capacitance and Potential

1
Chapter 18.2 Review Capacitance and Potential
2
1. A 5 µF capacitor is connected to a 12 volt
battery. What is the potential difference across
the plates of the capacitor when it is fully
charged?
3
12 volts, a charged capacitor has the same
potential difference across the plates as the
source.
4
2. What is the charge on one plate of the
capacitor in problem 1 when it is fully charged?
What is the net charge on the capacitor when it
is fully charged?
5
q VC q 12 x 5 60 µC or q 12(5 x 10-6) 6
x 10-5 C But the net charge on a capacitor is
always zero because the and plate have
charges of equal magnitude.
6
3. How much electrical potential energy is stored
in the capacitor in problem 1?
7
PEE ½ CV2 PEE ½ 5(12)2 PEE 360 µJ or, PEE
½ (5x 10-6)(12)2 PEE 3.6 x 10-4 J
8
4. If the plates are separated by a distance of
0.003 m, what is the strength of the electric
field across the plates?
9
Ed V E(0.003) 12 E 4000 N/C
10
5. A 9 µF capacitor is connected to a 9 volt
battery. What is the potential difference across
the plates of the capacitor when it is fully
charged?
11
9 volts, a charged capacitor has the same
potential difference across the plates as the
source.
12
6. What is the charge on one plate of the
capacitor in problem 5 when it is fully charged?
What is the net charge on the capacitor when it
is fully charged?
13
q VC q 9 x 9 81 µC or q 9(9 x 10-6) 8.1
x 10-5 C But the net charge on a capacitor is
always zero because the and plate have
charges of equal magnitude.
14
7. How much electrical potential energy is stored
in the capacitor in problem 5?
15
PEE ½ CV2 PEE ½ 9(9)2 PEE 364.5 µJ or, PEE
½ (9 x 10-6)(9)2 PEE 3.645 x 10-4 J
16
8. If the plates are separated by a distance of
0.005 m, what is the strength of the electric
field across the plates?
17
Ed V E(0.005) 9 E 1800 N/C
18
9. List four ways to increase the charge on a
capacitor.
19
1. Larger plates 2. Plates closer together 3.
Different dielectric 4. Increase voltage
20
10. What is the difference in electrical
potential energy and electric potential?
21
Electric potential is electric potential energy
divided by charge.
22
11. A proton is moved 0.5 m in the direction of
an electric field with a strength of 5000 N/C.
What is the change in electric potential?
23
Ed V 5000 x 0.5 V 2500 V V Moving in the
direction of the field is a decrease of
potential, so -2500 V V.
24
12. What is the change in electrical potential
energy of the proton in problem 11?
25
PEE qV PEE (1.6 x 10-19)(2500) PEE -4 x
10-16 J
26
13. An electron is moved 0.8 m in the direction
of an electric field with a strength of 3000 N/C.
What is the change in electric potential?
27
Ed V 3000 x 0.8 V 2400 V V Moving in the
direction of the field is a decrease of
potential, so -2400 V V.
28
14. What is the change in electrical potential
energy of the electron in problem 13?
29
PEE qV PEE (-1.6 x 10-19)(-2400) PEE 3.88
x 10-16 J The electron is being moved in the
direction it does not want to go. That is an
increase in potential energy.
30
15. A helium nucleus (composed of two protons and
two neutrons) is moved 24 cm in the direction of
an electric field with a strength of 9000 N/C.
What is the change in electric potential?
31
Ed V 9000 x 0.24 V 2160 V V Moving in the
direction of the field is a decrease of
potential, so -2160 V V.
32
16. What is the change in electrical potential
energy of the helium nucleus in problem 15?
33
PEE qV PEE (1.6 x 10-19)(2)(-2160) (There
are two protons in one helium nucleus.) PEE
-6.912 x 10-16
34
(No Transcript)
35
(No Transcript)