Question

A point charge of +1.0 µC is moved in the direction of an electric field, and it has a change in electric potential difference energy of 10.0 J. What was the change in electric potential difference? 1 µC = 10^-6 C

1.0 × 108 V

1.0 × 107 V

+1.0 × 107 V

+1.0 × 108 V

Answer 1

@Isaiah.Feynman

Answer 2

@ranga

Answer 3

The work done moving a charge in an electrical field is given by:\[W=q \Delta V\]where W is work (joules); q is the charge (Coulombs); and ∆V is the change in potential (volts).

Answer 4

Do you see, now, how to solve the problem?

Answer 5

What about this question
In the absence of friction, gravity, and all other external forces, what kind of speed will an object display if it travels from the bottom to the top of an inclined plane?

Answer 6

So the only force is the force pushing the object up an inclined plane?

Answer 7

So would the answer be 1 * 10^7?

Answer 8

@PsiSquared

Answer 9

Yes, the answer is B, 1*10^7V=10MV

Answer 10

Okay, thank you. what about this question?
A voltmeter reads 200.0 V between parallel plates that are 2.0 cm apart. What is the strength of the electric field between the two plates?
1.0 × 10^1 N/C

1.0 × 10^2 N/C

1.0 × 10^3 N/C

1.0 × 10^4 N/C

Answer 11

@psisquared

Answer 12

@IhteshamMalik

Answer 13

\[E = - \frac{ \Delta v }{ \Delta r }\]

Answer 14

E = 200/ 0.02 = \[1 \times 10^{3} \]

Answer 15

SO the answer is third one. and Volt / meter = Newton/coulomb

Answer 16

10,000 N per coulomb.

Answer 17

Oh, okay! that makes sense.

Answer 18

sense always makes sense... :D

Answer 19

lol, :) thanks for the help!

Answer 20

You're welcome bro.

Answer 21

what about this question?
If 4.0 × 1018 J of work done on a point charge changes its electric potential difference by 25.0 V, then what is the magnitude of the charge?
1.6 × 101^9 C

4.0 × 101^9 C

1.6 × 101^8 C

4.0 × 101^8 C

Answer 22

@IhteshamMalik

Answer 23

SO you are not gonna to sleep??? :D

Answer 24

\[q = \frac{ W }{ \Delta V }\]

Answer 25

@lmckenzie3

Answer 26

????

Answer 27

sorry, my interet connection went out for a while. I'm here. No, im not sleeping or going to go to sleep. haha

Answer 28

@IhteshamMalik

Answer 29

okay solve your question then with above formula...

Answer 30

Okay, so would th answer be 1.6 * 10^-19?

Answer 31

ooo how did you do this???

Answer 32

Q = W/V = \[q = \frac{ 4 \times10^{18} }{ 25 }\]

Answer 33

i did exactly like the formula above but came up with the answer 1.6 * 10^-19 on my calculator.

Answer 34

Did i wrote right value of work done???

Answer 35

and check your answers also

Answer 36

The answers i typed above are incorrect. I just now realized that. The choices are supposed to be

1.6 × 10^19 C

4.0 × 10^19 C

1.6 × 10^18 C

4.0 × 10^18 C

Answer 37

i have answer. \[1.6 \times 10^{17} C\] and i have no doubt about answer.

Answer 38

Check question also, i think \[W = 4 \times 10^{19}\]

Answer 39

hmm what you thinking??

Answer 40

Oh you are right thank you! can you help me with this other question?
For the electric field shown, which of the following statements describes an increase in voltage?
A positive charge moves along an equipotential line.

A positive charge crosses an equipotential line toward the source charge.

A negative charge crosses an equipotential line toward the source charge.

You cannot tell from the information given.

Answer 41

I know the answer is either A or B. I just am stuck between the two and dont know which one it is

Answer 42

Point B is true.

Answer 43

Because we have the equation \[ V = k \frac{ q }{ r }\]

Answer 44

So increase in distance from the source, causes decrease in potential.