Using the equation for Coulomb's Law, what happens to the force between two objects when you double the distance between them?
\large F=k\frac{q_1q_2}{r^2}
a
it quadruples
b
it doubles
c
it decreases by half
d
it decreases by 1/4
The correct answer is (d) it decreases by 1/4.
Coulomb's Law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
If the distance between the charges is doubled, the denominator of the fraction (r^2) is now four times larger than before. This means that the overall fraction is one-fourth of its original value. Therefore, the force between the charges decreases by a factor of 1/4.
To determine what happens to the force between two objects when you double the distance between them using Coulomb's Law, we can simply analyze the equation:
\large F=k\frac{q_1q_2}{r^2}
Where:
- F represents the force between the two objects
- k is the electrostatic constant
- q1 and q2 are the charges of the objects
- r is the distance between the objects
Now, let's consider what happens when we double the distance (r) between the two objects. We can express this change as:
\large r_{\text{new}} = 2r_{\text{old}}
Substituting this new distance into the equation, we get:
\large F_{\text{new}} = k\frac{q_1q_2}{(2r_{\text{old}})^2}
Simplifying further:
\large F_{\text{new}} = k\frac{q_1q_2}{4r_{\text{old}}^2}
Notice that the denominator now contains 4 times the value of the original distance (r_old). Therefore, when we double the distance between the objects, the force (F) decreases by a factor of 4. This means that the force decreases by 1/4 of its original value.
So, the correct answer is:
d) it decreases by 1/4
According to Coulomb's Law, the force between two objects is inversely proportional to the square of the distance between them.
So, if you double the distance between two objects, the force between them will decrease by a factor of 1/4.
Therefore, the correct answer is option D: it decreases by 1/4.