two charge 5NC and -2NC are placed at a point (5cm,0,0)and (23cm,0,0)in a region of a space where there is no other electric field.calculate the electrostatic potential energy of this charged system
k =9•10⁹ N•m²/C²
r=0.18 m
q₁= 5•10⁻⁹ C
q₂ = - 2•10⁻⁹ C
PE= k•q₁•q₂/r = ...
The electrostatic potential energy of a charged system can be calculated using the formula:
U = k * q1 * q2 / r
Where U is the potential energy, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
In this case, the charges are +5NC and -2NC, so q1 = 5NC and q2 = -2NC. The distance between the charges is the difference in their positions, which is 23cm - 5cm = 18cm = 0.18m (since 1cm = 0.01m).
Now we can substitute these values into the formula to calculate the potential energy:
U = (9 x 10^9 Nm^2/C^2) * (5NC) * (-2NC) / 0.18m
Simplifying:
U = -10 * (9 x 10^9 Nm^2/C^2) * NC^2 / 0.18m
U = -50 * (9 x 10^9 Nm^2/C^2) * NC^2 / m
U = -50 * (9 x 10^9 N) * (C^2 / m)
U = -450 x 10^9 N * C^2 / m
So, the electrostatic potential energy of this charged system is -450 x 10^9 N * C^2 / m. The negative sign indicates that the potential energy is negative, indicating an attractive force between the charges.