Finding potential difference given work and kinetic energy

[leolaw]

The work done by an external force to move a $$-8.50{\mu}C$$ charge from a point a to point b is $$15.0*10^{-4}J$$. If the charge was started from rest and had $$4.82 * 10^{-4}J$$ of kinetic energy when it reached point b, what must be the potential difference between a and b.

I am not sure if this is suppose to be a trick question or not, but if the final kinetic energy is 4.82 * 10^ -4, then the initial potential difference has to be 4.82 * 10^-4 right?

 

[vsage]

Consider the units of electric potention (V or J/C) and work (J)

 

[dextercioby]

What is the unit of potential difference...?

Daniel.

 

[leolaw]

so
$${\Delta} KE = -{\Delta}PE}$$
$$\frac{1}{2}mv^2 = -qV_{ba}$$
$$4.82*10^{-4}J=8.5*10^{-6} V_{ba}$$
So V_ba = 5.6 * 10 V

But when I use
$$W = -qV_{ba}$$
$$15 * 10^{-4} = 8.5 * 10^{-6} V_{ba}$$
$$V_{ba} = 1.76 * 10 ^2$$

I am kind of lost here, can somebody help

 

[xanthym]

You're not properly using the energy information provided. What's the relationship between the 2 energy values given in the problem?

~~

 

[leolaw]

$$4.82 * 10^{-4}J$$, which is the kinetic energy

 

[xanthym]

What about:

$$Work_{ext} = \Delta K.E. + \Delta P.E.$$


~~

 

[xanthym]

HINT:

$$:(1): \ \ \ \ Work_{ext} = \Delta K.E. + \Delta P.E.$$

$$:(2): \ \ \ \ (15x10^{-4} Joules) \ \ = \ \ (4.82x10^{-4} Joules) \ \ + \ \ (Charge)*(PotentialDifference)$$

Do you see the technique?


~~

 

[leolaw]

I have the answer -1.19 x 10 ^ -2 V

 

[dextercioby]

I've said that b4.It's the OP-s job to do the "dirty" work.Namely the simple calculations...We assume that the person coming in for advice/help has minimum knowledge of arithmetics...

Daniel.

P.S.Your result is 10.000 times smaler...
P.P.S.xanthym,the latex code for multiplication is either "\times" or "\cdot" or leaving a space...

 

[leolaw]

What about:

$$Work_{ext} = \Delta K.E. + \Delta P.E.$$


~~

I didnt know this formula before because I started from the middle of the book (since that i am taking second half of the course), so i skipped all the Mechnical energy stuff

But once i have this, it makes everything easier

 

[leolaw]

I've said that b4.It's the OP-s job to do the "dirty" work.Namely the simple calculations...We assume that the person coming in for advice/help has minimum knowledge of arithmetics...

Daniel.

P.S.Your result is 10.000 times smaler...
P.P.S.xanthym,the latex code for multiplication is either "\times" or "\cdot" or leaving a space...

you mean it is -1.2 x 10 ^3?

 

[xanthym]

LEOLAW --
Don't guess. Repeat the calculation. (10^3 is still wrong.)

 

[leolaw]

HINT:

$$:(1): \ \ \ \ Work_{ext} = \Delta K.E. + \Delta P.E.$$

$$:(2): \ \ \ \ (15x10^{-4} Joules) \ \ = \ \ (4.82x10^{-4} Joules) \ \ + \ \ (Charge)*(PotentialDifference)$$

Do you see the technique?


~~

$$(15*10^{-4} Joules) = (4.82x10^{-4} Joules) + (Charge)*(PotentialDifference)$$

$$1.018 * 10 ^{-3} = 8.5*10^{-6} * (PotentialDifference)$$

and i still get the result -1.19 x 10 ^2

now I get the idea... but not the answer?/

 

[xanthym]

CORRECT!

 

[leolaw]

CORRECT!

finally..wholala