Two charges are located along the x-axis. One
has a charge of 6.2 µC, and the second has a
charge of −3.1 µC.
The Coulomb constant is ke =
8.98755 × 10
9
N m2
/C
2
The acceleration of .
gravity is 9.81 m/s
2
.
If the electrical potential energy associated
with the pair of charges is −3.5×10
−2
J, what
is the distance between the charges?
Answer in units of m
To find the distance between the charges, we need to use the equation for electrical potential energy:
U = k * (q1 * q2) / r
Where:
U = electrical potential energy
k = Coulomb constant (8.98755 × 10^9 N m^2/C^2)
q1 and q2 = charges of the two particles
r = distance between the charges
Given that the electrical potential energy is -3.5 × 10^-2 J and the charges are 6.2 µC and -3.1 µC, we can substitute these values into the equation:
-3.5 × 10^-2 J = (8.98755 × 10^9 N m^2/C^2) * (6.2 µC * (-3.1 µC)) / r
Now, let's solve for the distance (r):
r = (8.98755 × 10^9 N m^2/C^2) * (6.2 µC * (-3.1 µC)) / (-3.5 × 10^-2 J)
First, let's calculate the numerator:
(6.2 µC * (-3.1 µC)) = -19.22 µC^2
Now, let's substitute this value into the equation:
r = (8.98755 × 10^9 N m^2/C^2) * (-19.22 µC^2) / (-3.5 × 10^-2 J)
Next, we simplify the equation:
r = (-180.68231 N m^2) / (-3.5 × 10^-2 J)
To get the final answer, we divide the numerator by the denominator and convert the units:
r = 5159.494 m^2/s^2
Finally, taking the square root to get the distance:
r = √5159.494
r ≈ 71.83 m
Therefore, the distance between the charges is approximately 71.83 m.