MHB Electrostatic Force: Effects of Distance

Click For Summary
Electrostatic force, like gravitational force, follows an inverse square law, meaning that if the distance between two charged objects is doubled, the force between them decreases by a factor of one-fourth. This relationship is described by Coulomb's Law, which states that the electrostatic force is proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula for electrostatic force is \(F_e = k_e\frac{|q_1q_2|}{d^2}\). Thus, both gravitational and electrostatic forces exhibit similar behavior regarding distance. Understanding this principle is crucial for applications in physics and engineering.
Dustinsfl
Messages
2,217
Reaction score
5
How does electrostatic force vary between two objects if the distance is doubled?

I know with gravitational force as the distance doubles the force decreases by \(\frac{1}{4}\).
 
Mathematics news on Phys.org
dwsmith said:
How does electrostatic force vary between two objects if the distance is doubled?

I know with gravitational force as the distance doubles the force decreases by \(\frac{1}{4}\).

Gravitational and Electrostatic force fields are both inverse square fields. In particular for gravity (known as Newton's Law of Universal Gravitation),

\[F_g = G\frac{m_1m_2}{d^2}\]

where $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of objects and $d$ is the distance between these two objects.

Likewise for electrostatic force (known as Coulomb's Law),

\[F_e = k_e\frac{|q_1q_2|}{d^2}\]

where $k_e$ is Coulomb's constant, $q_1$ and $q_2$ are the signed charges of the particles, and $d$ is the distance between these two particles.

So if the distance between any two particles/objects is doubled in either case, the force decreases by a factor of $\dfrac{1}{4}$.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
14
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
16
Views
1K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
690
  • · Replies 12 ·
Replies
12
Views
1K
  • · Replies 11 ·
Replies
11
Views
5K