Peoples Physics Concepts

12.3. Voltage http://www.ck12.org

Example 1

You have a negative charge of unknown value and a positive charge of magnitudeq 1 and massm. After fixing the

negative charge in place, you place the positive charge a distanceriaway from the negative charge and then release

it. If the speed of the positive charge when it is a distancerfaway from the negative charge isv, what was the

magnitude of the negative charge in terms of the given values?

Solution

There are multiple ways to do this problem, we will solve it using conservation of energy and the change in voltage

to determine the magnitude of the negative charge. When working through this problem, we’ll call the positive

chargeq 1 and the negative chargeq 2. To start we’ll say that the charge had zero potential energy when it was .5m

from the negative charge; this will help us as we work through the problem. Using this assertion, we will apply

conservation of energy to the positive charge.

∆Ue=KEf start with conservation of energy

q 1 ∆V=

1

2

mv^2 substitute the equations for each energy term

∆V=

mv^2

2 q 1

solve for V

Now, since we know the voltage difference, we will express it using the equation for voltage at a certain distance

from a point charge.

∆V=

kq 2

∆r

start with the equation for voltage at a certain distance

∆V=

kq 2

rf

−

kq 2

ri

express the change in radius in terms of the initial and final radius of the positive charge

∆V=kq 2 (

1

rf

−

1

ri

) factor the equation

mv^2

2 q 1

=kq 2 (

1

rf

−

1

ri

) substitute in the value from the first step

q 2 =

mv^2

2 q 1 k(r^1 f−^1 ri)

solve forq 2

Watch this Explanation

MEDIA

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