Potential difference and Kinetic energy

In summary, the object gained kinetic energy from an electric field, which can be calculated using the equation KE = (1/2)mv^2. The amount of energy gained can be found by multiplying the object's charge (20uC) by the potential difference (V). By using this information and setting the gain in kinetic energy equal to the loss of electrical potential energy, we can solve for the magnitude of the potential difference.
  • #1
donjt81
71
0
A 4.0g object carries a charge of 20uC. The object is accelerated from rest through a potential difference, and afterward the ball is moving at 2.0 m/s. What is the magnitude of the potential difference?

I know i have to use the law of conservation

Potential_initial + KE_initial = Potential_final + KE final
Potential_initial + 0 = 0 + KE final

but I don't know what to do after this. can some one give me any pointers?
 
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  • #2
The object gained some amount of kinetic energy, which can found with

[itex]
KE = \frac{1}
{2}mv^2
[/itex]

This energy had to come from somewhere; in this case, it came from an electric field.

The amount of energy gained by an object of charge q, moving through a potential difference of V volts is:

[itex]
E = qV
[/itex]

The easiest way to realize this is simply to look at the units. The "volt" is exactly equal to one joule per coulomb. Look at how the factors cancel if you use this definition of the volt:

[itex]
{\text{20}}\,\mu {\text{C}} \cdot V\frac{{{\text{joules}}}}
{{{\text{coulomb}}}}
[/itex]

The coulombs cancel, leaving you with just joules.

All you need do is to solve for V, such that the gain in kinetic energy is equal to the loss of electrical potentail energy:

[itex]
\frac{1}
{2}mv^2 = qV
[/itex]

- Warren
 
  • #3



Sure, I can provide some pointers. First, let's define some variables:
- m = mass of the object (4.0g)
- q = charge on the object (20uC)
- v = final velocity of the object (2.0 m/s)
- V = potential difference

Now, let's use the equation for kinetic energy:
KE = 1/2 * m * v^2

We can plug in our known values:
KE = 1/2 * (0.004 kg) * (2.0 m/s)^2 = 0.004 J

Next, let's use the equation for potential energy:
PE = q * V

We can rearrange this to solve for V:
V = PE / q

Since we know that the potential difference is the same at the beginning and end of the acceleration, we can say that the initial potential energy is equal to the final potential energy. Therefore, we can say that:
PE_initial = PE_final

Substituting in our values:
q * V = q * V
(20 x 10^-6 C) * V = (20 x 10^-6 C) * V

Now we can solve for V:
V = KE / q = 0.004 J / (20 x 10^-6 C) = 200 V

So, the magnitude of the potential difference is 200 volts. I hope this helps!
 

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electrical potential energy between two points in a circuit. It is measured in volts (V) and is the driving force that causes electric charges to move through a circuit.

2. How is potential difference related to kinetic energy?

Potential difference and kinetic energy are both forms of energy. Potential difference is the energy stored in an electric field, while kinetic energy is the energy of motion. When an electric charge moves through a circuit, it converts potential energy into kinetic energy as it gains speed.

3. What is the formula for calculating potential difference?

The formula for calculating potential difference is V = W/Q, where V is the potential difference in volts, W is the work done on the charge in joules (J), and Q is the charge in coulombs (C). This formula can also be written as V = IR, where I is the current in amperes (A) and R is the resistance in ohms (Ω).

4. How does potential difference affect the speed of charges in a circuit?

The greater the potential difference, the greater the speed of charges in a circuit. This is because potential difference is the driving force that causes charges to move, and the higher the potential difference, the more work is done on the charges, increasing their kinetic energy and speed.

5. Can potential difference be negative?

Yes, potential difference can be negative. This occurs when the direction of the electric field is opposite to the direction of the charge's motion. In this case, the potential energy of the charge decreases, resulting in a negative potential difference.

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