Question
Two equal and opposite charges separated by a distance, d, experience an attractive force, F. If each charge is doubled in magnitude, what should be the new distance between them so that the force remains unchanged?
Responses
A 2d
B 4d
C d/2
D d/4
Responses
A 2d
B 4d
C d/2
D d/4
Answers
GPT-4o mini
To find out how the distance should change while keeping the force the same after doubling the charges, we can use Coulomb's law, which states:
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
where \( F \) is the force between the charges, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
Initially, let the charges be \( q \) and \( -q \), and the distance between them be \( d \). The initial force is:
\[
F = k \frac{|q \cdot (-q)|}{d^2} = k \frac{q^2}{d^2}
\]
Now, when each charge is doubled, they become \( 2q \) and \( -2q \). Let the new distance between them be \( r \). The new force \( F' \) will then be:
\[
F' = k \frac{|2q \cdot (-2q)|}{r^2} = k \frac{4q^2}{r^2}
\]
We want the new force to equal the original force:
\[
F' = F
\]
Substituting the expressions we have, we get:
\[
k \frac{4q^2}{r^2} = k \frac{q^2}{d^2}
\]
We can cancel \( k \) and \( q^2 \) from both sides (assuming \( q \neq 0 \)):
\[
\frac{4}{r^2} = \frac{1}{d^2}
\]
Cross-multiplying gives:
\[
4d^2 = r^2
\]
Taking the square root of both sides:
\[
r = 2d
\]
Thus, the new distance \( r \) should be \( 2d \) for the force to remain unchanged.
The correct response is:
**A) 2d**
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
where \( F \) is the force between the charges, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
Initially, let the charges be \( q \) and \( -q \), and the distance between them be \( d \). The initial force is:
\[
F = k \frac{|q \cdot (-q)|}{d^2} = k \frac{q^2}{d^2}
\]
Now, when each charge is doubled, they become \( 2q \) and \( -2q \). Let the new distance between them be \( r \). The new force \( F' \) will then be:
\[
F' = k \frac{|2q \cdot (-2q)|}{r^2} = k \frac{4q^2}{r^2}
\]
We want the new force to equal the original force:
\[
F' = F
\]
Substituting the expressions we have, we get:
\[
k \frac{4q^2}{r^2} = k \frac{q^2}{d^2}
\]
We can cancel \( k \) and \( q^2 \) from both sides (assuming \( q \neq 0 \)):
\[
\frac{4}{r^2} = \frac{1}{d^2}
\]
Cross-multiplying gives:
\[
4d^2 = r^2
\]
Taking the square root of both sides:
\[
r = 2d
\]
Thus, the new distance \( r \) should be \( 2d \) for the force to remain unchanged.
The correct response is:
**A) 2d**