FHSST Physics/Electrostatics/Electrostatic Force

Electrostatic Force

As we now know, charged objects exert a force on one another. If the charges are at rest then this force between them is known as the electrostatic force. An interesting characteristic of the electrostatic force is that it can be either attractive or repulsive, unlike the gravitational force which is only ever attractive. The relative charges on the two objects is what determines whether the force between the charged objects is attractive or repulsive. If the objects have opposite charges they attract each other, while if their charges are similar they repel each other (e.g. two metal balls which are negatively charged will repel each other, while a positively charged ball and negatively charged ball will attract one another).

It is this force that determines the arrangement of charge on the surface of conductors. When we place a charge on a spherical conductor the repulsive forces between the individual like charges cause them to spread uniformly over the surface of the sphere. However, for conductors with non-regular shape there is a concentration of charge near the point or points of the object.

This collection of charge can actually allow charge to leak off the conductor if the point is sharp enough. It is for this reason that buildings often have a lightning rod on the roof to remove any charge the building has collected. This minimises the possibility of the building being struck by lightning.

This spreading out of charge would not occur if we were to place the charge on an insulator since charge cannot move in insulators.

Coulomb's Law

The behaviour of the electrostatic force was studied in detail by Charles Coulomb around 1784. Through his observations he was able to show that the electrostatic force between any point two point charges is directly proportional to their magnitude and inversely proportional to the square of the distance between the charges. Thus if F is the force, Q1 and Q2 are the charges and R is their distance apart. then the law is summarized as

                F=(Q1Q2)/(R^2) 
                          

where Q₁ is the charge on the one point-like object, Q₂ is the charge on the second, and r is the distance between the two. Remember her F is the mutual force that acts on both the charges exerted by one another as in Newtons third law of motion.

The magnitude of the electrostatic force between two point-like charges is given by Coulomb's Law:

${\begin{matrix}F=k{\frac {Q_{1}Q_{2}}{r^{2}}}\end{matrix}}$

(12.1)

and the proportionality constant k is called the electrostatic constant. We will use the value

${\begin{matrix}k=8.99\times 10^{9}{\rm {N\cdot m^{2}/C^{2}.}}\end{matrix}}$

The value of the electrostatic constant is known to a very high precision (9 decimal places). Not many physical constants are known to as high a degree of accuracy as k.

Aside: Notice how similar Coulomb's Law is to the form of Newton's Universal Law of Gravitation between two point-like particles:

${\begin{matrix}F_{G}=G{\frac {m_{1}m_{2}}{r^{2}}},\end{matrix}}$

where m₁ and m₂ are the masses of the two particles, r is the distance between them, and G is the gravitational constant. It is very interesting that Coulomb's Law has been shown to be correct no matter how small the distance, nor how large the charge: for example it still applies inside the atom (over distances smaller than $10^{-10}{\rm {m}}$

Let's run through a simple example of electrostatic forces.

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Worked Example 59 Coulomb's Law I

Question: Two point-like charges carrying charges of +3×10⁻⁹ C and −5×10⁻⁹ C are 2 m apart. Determine the magnitude of the force between them and state whether it is attractive or repulsive.

Answer:

Step 1 :

First draw the situation:

Step 2 :

Is everything in the correct units? Yes, charges are in coulombs [C] and distances in meters [m].

Step 3 :

Determine the magnitude of the force: Using Coulomb's Law we have

${\begin{matrix}F&=&k{\frac {Q_{1}Q_{2}}{r^{2}}}\\&=&(8.99\times 10^{9}{\rm {N\cdot m^{2}/C^{2}}}){\frac {(+3\times 10^{-9}{\rm {C}})(-5\times 10^{-9}{\rm {C}})}{(2{\rm {m}})^{2}}}\\&=&-{\frac {(8.99)(3)(5)}{4}}(10^{-9}){\frac {(N\cdot m^{2}/C^{2}\!\!\!/)(C^{2})}{m^{2}}}\\&=&-3.37\times 10^{-8}{\rm {N}}\end{matrix}}$

Thus the magnitude of the force is $3.37\times 10^{-8}{\rm {N}}$. The minus sign is a result of the two point charges having opposite signs.

Step 4 :

Is the force attractive or repulsive? Well, since the two charges are oppositely charged, the force is attractive. We can also conclude this from the fact that Coulomb's Law gives a negative value for the force.

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Next is another example that demonstrates the difference in magnitude between the gravitational force and the electrostatic force.

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Worked Example 60 Coulomb's Law II

Question: Determine the electrostatic force and gravitational force between two electrons 1Aring apart (i.e. the forces felt inside an atom) Answer: Å Step 1 :

First draw the situation:

Riaan Note: this image is faulty, recreate

Step 2 :

Get everything into S.I. units: The charge on an electron is $-1.60\times 10^{-19}{\rm {C}}$, the mass of an electron is $9.11\times 10^{-31}{\rm {kg}}$, and 1 Å = $1\times 10^{-10}{\rm {m}}$

Step 3 :

Calculate the electrostatic force using Coulomb's Law:

${\begin{matrix}F_{E}&=&k{\frac {Q_{1}Q_{2}}{r^{2}}}=k{\frac {e\cdot e}{1\mathrm {\AA} ^{2}}}\\&=&(8.99\times 10^{9}{\rm {N\cdot m^{2}/C^{2}){\frac {(-1.60\times 10^{-19}{\rm {C}})(-1.60\times 10^{-19}{\rm {C)}}}{(10^{-10}{\rm {m)^{2}}}}}}}\\&=&2.30\times 10^{-8}{\rm {N}}\end{matrix}}$

Hence the magnitude of the electrostatic force between the electrons is $2.30\times 10^{-8}{\rm {N}}$. (Note that the electrons carry like charge and from this we know the force must be repulsive. Another way to see this is that the force is positive and thus repulsive.)

Step 4 :

Calculate the gravitational force:

${\begin{matrix}F_{E}&=&G{\frac {m_{1}m_{2}}{r^{2}}}=G{\frac {m_{e}\cdot m_{e}}{(1\mathrm {\AA} )^{2}}}\\&=&(6.67\times 10^{-11}{\rm {N\cdot m^{2}/kg^{2}}}){\frac {(9.11\times 10^{-31}{\rm {C}})(9.11\times 10^{-31}{\rm {kg}})}{(10^{-10}{\rm {m}})^{2}}}\\&=&5.54\times 10^{-51}{\rm {N}}\end{matrix}}$

The magnitude of the gravitational force between the electrons is $5.54\times 10^{-51}{\rm {N}}$

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Notice that the gravitational force between the electrons is much smaller than the electrostatic force. For this reason, the gravitational force is usually neglected when determining the force between two charged objects.

We mentioned above that charge placed on a spherical conductor spreads evenly along the surface. As a result, if we are far enough from the charged sphere, electrostatically, it behaves as a point-like charge. Thus we can treat spherical conductors (e.g. metallic balls) as point-like charges, with all the charge acting at the centre.

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Worked Example 61 Coulomb's Law: Challenge Question

Question: In the picture below, X is a small negatively charged sphere with a mass of 10 kg. It is suspended from the roof by an insulating rope which makes an angle of 60^o with the roof. Y is a small positively charged sphere which has the same magnitude of charge as X. Y is fixed to the wall by means of an insulating bracket. Assuming the system is in equilibrium, what is the magnitude of the charge on X?

Answer: How are we going to determine the charge on X? Well, if we know the force between X and Y we can use Coulomb's Law to determine their charges as we know the distance between them. So, firstly, we need to determine the magnitude of the electrostatic force between X and Y.

Step 1 :

Is everything in S.I. units? The distance between X and Y is $50{\rm {cm}}=0.5{\rm {m}}$, and the mass of X is 10 kg.

Step 2 :

Draw the forces on X (with directions) and label.

Step 3 :

Determine the magnitude of the electrostatic force (F_E). Since nothing is moving (system is in equilibrium) the vertical and horizontal components of the forces must cancel. Thus

${\begin{matrix}F_{E}=T\sin(60^{o})\qquad \qquad F_{g}=T\sin(60^{o}).\end{matrix}}$

The only force we know is the gravitational force F_g = mg. Now we can calculate the magnitude of T from above:

${\begin{matrix}T={\frac {F_{g}}{\sin(60^{o})}}={\frac {(10{\rm {kg}})(10{\rm {m/s^{2})}}}{\sin(60^{o})}}=1155{\rm {N}}.\end{matrix}}$

Which means that F_E is:

${\begin{matrix}F_{E}=T\cos(60^{o})=1154{\rm {N\cdot \cos(60^{o})=577.5{\rm {N}}}}\end{matrix}}$

Step 4 :

Now that we know the magnitude of the electrostatic force between X and Y, we can calculate their charges using Coulomb's Law. Don't forget that the magnitudes of the charges on X and Y are the same: $|Q_{\rm {X}}|=|Q_{\rm {Y}}|$. The magnitude of the electrostatic force is

${\begin{matrix}F_{E}&=&k{\frac {|Q_{\rm {X}}Q_{\rm {Y}}|}{r^{2}}}=k{\frac {Q_{\rm {X}}^{2}}{r^{2}}}\\|Q_{\rm {X}}|&=&{\sqrt {\frac {F_{E}r^{2}}{k}}}\\&=&{\sqrt {\frac {(577.5{\rm {N}})(0.5{\rm {m}})^{2}}{8.99\times 10^{9}{\rm {N\cdot m^{2}/C^{2}}}}}}\\&=&1.27\times 10^{-4}{\rm {C}}\end{matrix}}$

Thus the charge on X is $-1.27\times 10^{-4}{\rm {C}}$