# Problem involving Coulomb's law

1. Nov 17, 2015

### 1729

1. The problem statement, all variables and given/known data
A big metal sphere has radius $r$ and charge $Q$. The distance between a certain point charge and the surface of the sphere is equal to $s$. The magnitude of the force exerted on the point charge is inversely proportional to:

A) $s^2$
B) $r^2$
C) $(s+r)^2$ (correct answer)
D) $s^2+r^2$

(Source: Fysica Vandaag 5.2, Uitgeverij Pelckmans)

2. Relevant equations
Coulomb's law

3. The attempt at a solution
Let $d$ be the distance between the two charges.
$F=k\frac{|Q||Q'|}{d^2}=\frac{\mathrm{constant}}{d^2}$
This means the constant of proportionality is equal to $d^2$.

Why is $d=s+r$? I would expect it to be equal to $s$ since the charge in a sphere is the same everywhere.

Last edited: Nov 17, 2015
2. Nov 17, 2015

### SammyS

Staff Emeritus
If the sphere is uniformly charged, then at any point external to the sphere, the electric field produced by the sphere is the same as if all of the charge would reside at the center of the sphere.

3. Nov 17, 2015

### Staff: Mentor

Consider two extreme situations. First, take r = 1 mm, s = 10 cm. At that distance, the metal sphere looks almost like a point charge, and there is no significant difference between the interaction of the point charge with the part of the sphere that is closest, and the part that is farthest away.

Second, take r = 1 m and s = 10 cm. Since, as you correctly state, the charge Q is uniformly distributed, the point charge will feel a greater force from the part of the charge that is closest compared to that on the other side of the sphere, which is now 2 m away.

So, it can't depend only on s.

The reason why it is r + s has just been given by Sammy.

4. Nov 17, 2015

### 1729

Thank you for your answers! I see the error of my ways now -- your argument is convincing, DrClaude!

At first I was skeptical about the relation between the electric field property Sammy mentioned and Coulomb's law. Luckily, it then struck me that one can also define Coulomb's law in terms of the electric field force.

$F=k\frac{|Q||Q'|}{d^2}=|Q|\cdot E$
By the property Sammy mentioned, we find that the distance can be rewritten as $s+r$, and the result follows.

(In retrospect the electric field was a good way to think about this problem, but I didn't think of using it since the concept was introduced in the following chapter)

5. Nov 17, 2015

### Ray Vickson

Interestingly, this works only for an inverse-square force law. If the force between two charges at distance $r$ were
$$F = k \frac{q_1 q_2}{r^p}$$
for any $p\neq 2$, then the nice property cited above would no longer hold. There is something almost magical about the special value $p = 2$.

6. Nov 18, 2015

### 1729

That's beyond the scope of the course I'm taking, but interesting nevertheless. Could you expand on why that happens?

7. Nov 18, 2015

### azizlwl

If the distance far greater than the radius of the sphere, then we can assume it a point of charge. If not, there are many points of charges on the sphere to consider. Summing them all will show that it will be equalled to point of total charges at the center.