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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Monday, June 7, 2010

### AP Physics C - The Concept of Potential Energy - How it Simplifies Problem Solving

The concept of potential energy such as electrostatic potential energy, gravitational potential energy, elastic potential energy and the like usually makes seemingly difficult problems in physics simple to solve. Let us consider an example in electrostatics: Three point positive charges Q1, Q2 and Q3 are arranged equidistant R from the origin O as shown in the adjoining figure. Another point positive charge q of mass m, initially at rest, is released from the origin O. Derive an expression for the velocity of the point charge when it is far away from the origin.

This question can be made simpler if there is only a single charge Q1 instead of three charges Q1, Q2 and Q3. The complexity of the question can be increased further if the three charges Q1, Q2 and Q3 are at unequal distances from the origin.

Well, let us come to the question as it is. Since you are asked to determine the velocity of the charge q, you may be tempted to think of the electric field and the force (in fact, the resultant force) acting on the charge and the acceleration it produces. Equations you have often used in kinematics also may come to your mind. But the force and the accelearation are variable in this case and you realize that the method you plan to use does not seem to be workable. (A bright student will not have confusions of this sort and he will proceed in the right direction).

Perhaps you might have started thinking in terms of the electric potential and the potential energy if you were asked to determine the kinetic energy of the charge q when it is far away from the origin O. Electrostatic potential is a scalar quantity (unlike electric field, which is a vector quantity) and you can easily manipulate it. In the present problem, you can easily find the net electrostatic potential (due to the charges Q1, Q2 and Q3) at the origin and hence the electrostatic potential energy of the charge q. The electrostatic potential energy at infinity (far away from the origin) is zero and so the change in the potential energy of the charge q is equal to its potential energy at the origin O. By equating this to ½ mv2, you get the required velocity v of the point charge when it is far away from the origin O.

Here is how you will proceed:

The electric potential V at the origin (due to the charges Q1, Q2 and Q3) is given by

V = (1/4πε0)(Q1/R + Q2/R + Q3/R)

Or, V = (1/4πε0)[(Q1+ Q2+ Q3)/R]

[The electrostatic potential at any point is the work done by an external agency to bring a unit positive charge from infinity (infinite distance) to the point. Therefore, the electrostatic potential V at a point distant r from a point positive charge Q is given by

V =r [(1/4πε0)(Q/r2)]dr

The quantity inside the square bracket is the electrostatic force acting on the unit positive charge. When the charge Q is positive, the electric field produced by it and the displacement dr of the unit positive charge are opposite in direction and this is why the sign of the expression for V is negative. The above integral gives

V = (1/4πε0)(Q/r)

This shows that the potential energy is zero when r = ∞].

The electrostatic potential energy of the charge q when it is at the origin is (1/4πε0)[(Q1+ Q2+ Q3)q/R].

The loss of potential energy when the charge q moves from the origin O to a point far away from the origin is (1/4πε0)[(Q1+ Q2+ Q3)q/R] since the potential energy at infinity is zero.

The charge q gains an equivalent kinetic energy ½ mv2 and hence we have

½ mv2 = (1/4πε0)[(Q1+ Q2+ Q3)q/R]

The velocity v of the point charge when it is far away from the origin O is therefore given by

v = [(1/2πε0)(Q1+ Q2+ Q3)q/(mR)]1/2

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Now consider a simple question from gravitation:

An object of mass m is located at a point P very far away from the moon. The gravitational field of the moon is negligible at the point P and the object is initially at rest. The object is given a gentle push and it moves towards the moon. Determine the speed with which the object will strike the moon’s surface. Assume that the mass and radius of the moon are M and R respectively and the moon’s gravity alone influences the motion of the object.

You can apply the concept of gravitational potential energy to obtain the answer easily as given below:

Initial gravitational potential energy U1 of the object (at infinite distance from the centre of the moon, which we take as the origin) = 0.

[We have U = GMm/r where G is gravitational constant. With distance r = ∞, U = 0]

Final gravitational potential energy U2 of the object at the moon’s surface (at distance R) is given by

U2 = GMm/R

The loss of gravitational potential energy = U1 U2 = 0 – (– GMm/R) = GMm/R

This must be equal to the gain in kinetic energy ½ mv2 where v is the speed with which the object hits the moon’s surface.

Therefore, ½ mv2 = GMm/R, from which v = √(2GM/R)

Note that this is the expression for escape velocity from the surface of the moon. An object at infinity, gently pushed from rest must hit the surface (of the moon or any heavenly body) with the surface value of escape velocity since a body projected from the surface with escape velocity will reach infinite distance before coming to rest.

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The above question can almost equally well can be worked out beginning with the concept of force as follows

The gravitational force on the object of mass m at distance r = GMm/r2

Work done dW by the gravitational field in moving the object through a small distance dr along the direction of force is given by

dW = (GMm/r2)dr

We should have put the gravitational force as – GMm/R2 since it is directed opposite to the direction of increase of r. Our displacement is from infinity to R and hence dr too is negative. The work dW is indeed positive.

The total work W done by the gravitational field is given by

W = R(GMm/r2)dr = GMm/R

This is equal to the gain in kinetic energy ½ mv2. Therefore, ½ mv2 = GMm/R, from which v = √(2GM/R)