Charge, distance, E, and work

[combatwombat]

How much work is required to move a positively charged particle 15 cm if it is in an electric field of 10 N/C and the charge of the particle is 8 C?

A. 0.8 J
B. 8 J
C. 12 J
D. 1200 J

(From EK Physics, #146)

---

Members don't see this ad.

 

[Phlame217]

Well if you cant figure out the forumula, follow the units!

J = N*M
So you have N/C C and cm
Convert to appropriate units (ie cm - M) and figure out how to cancel units to get the correct units

Soooo N/C * C * M = N*M = J
Plug in numbers
(.15)(10)(8) = 1.5*8 = 12

Answer is 12 J

 

[combatwombat]

C is what I thought. However, the book says it's B. Their explanation: "The forces are conservative so if we turn the picture 90 degrees, this is just like gravity, mgh: the vertical distance h, and not the horizontal distance, is what matters. Similarly, in the question, only the distance against the electric field matters. The work required is force times distance parallel to the field, or Eqd."

 

[FuSoYa]

The issue is that 15cm is the entire path, but just like climbing a mountain, it doesn't matter what winding path you take. The only thing that matters is the difference between the start and end points, which in this problem is only 10cm. So the math is correct, but the numbers are not.

(.10)*(10)*(8) = 8 J

 

[physics junkie]

The issue is that 15cm is the entire path, but just like climbing a mountain, it doesn't matter what winding path you take. The only thing that matters is the difference between the start and end points, which in this problem is only 10cm. So the math is correct, but the numbers are not.

(.10)*(10)*(8) = 8 J

Fuyosa is correct, Work is only counted when moving in the direction of the electric field. Moving perpendicular to the direction of a field(such as in a circular fashion around the origin of the field) will not be counted as work.

 

[muchomaas]

Sorry, but where does it state in the problem that the distance from start to end is 10 cm?

 

[FuSoYa]

If you have the book, there's a diagram, which shows 10cm.