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 2d2 d
B 4d4 d
C d/2d/ 2
D d/4
Responses
A 2d2 d
B 4d4 d
C d/2d/ 2
D d/4
Answers
GPT-4o mini
We can use Coulomb's law to solve this problem. According to Coulomb's law, the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by:
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
where \( k \) is Coulomb's constant.
Initially, let the charges be \( q \) and \( -q \) (equal and opposite), and they are separated by a distance \( d \). The initial force \( F \) is:
\[
F = k \frac{|q \cdot -q|}{d^2} = k \frac{q^2}{d^2}
\]
Now, if each charge is doubled in magnitude, the charges become \( 2q \) and \( -2q \). Let the new distance between them be \( r \). The new force \( F' \) will be:
\[
F' = k \frac{|2q \cdot -2q|}{r^2} = k \frac{4q^2}{r^2}
\]
We want the new force \( F' \) to remain unchanged, that is, \( F' = F \). Setting the equations equal gives:
\[
k \frac{4q^2}{r^2} = k \frac{q^2}{d^2}
\]
We can cancel \( k \) and \( q^2 \) (assuming \( q \neq 0 \)) from both sides:
\[
\frac{4}{r^2} = \frac{1}{d^2}
\]
Cross-multiplying yields:
\[
4d^2 = r^2
\]
Taking the square root of both sides provides:
\[
r = 2d
\]
Thus, the new distance \( r \) between the charges, so that the force remains unchanged, is \( 2d \).
Therefore, the correct response is:
**A: 2d**
\[
F = k \frac{|q_1 q_2|}{r^2}
\]
where \( k \) is Coulomb's constant.
Initially, let the charges be \( q \) and \( -q \) (equal and opposite), and they are separated by a distance \( d \). The initial force \( F \) is:
\[
F = k \frac{|q \cdot -q|}{d^2} = k \frac{q^2}{d^2}
\]
Now, if each charge is doubled in magnitude, the charges become \( 2q \) and \( -2q \). Let the new distance between them be \( r \). The new force \( F' \) will be:
\[
F' = k \frac{|2q \cdot -2q|}{r^2} = k \frac{4q^2}{r^2}
\]
We want the new force \( F' \) to remain unchanged, that is, \( F' = F \). Setting the equations equal gives:
\[
k \frac{4q^2}{r^2} = k \frac{q^2}{d^2}
\]
We can cancel \( k \) and \( q^2 \) (assuming \( q \neq 0 \)) from both sides:
\[
\frac{4}{r^2} = \frac{1}{d^2}
\]
Cross-multiplying yields:
\[
4d^2 = r^2
\]
Taking the square root of both sides provides:
\[
r = 2d
\]
Thus, the new distance \( r \) between the charges, so that the force remains unchanged, is \( 2d \).
Therefore, the correct response is:
**A: 2d**