Two identical point charges are fixed to diagonally opposite corners of a square that is 1.5 m on a side. Each charge is +2.0 µC. How much work is done by the electric force as one of the charges moves to an empty corner?
I have W(fe)= -EPE=-q[V(f)-V(i)]. I can't figure out what to do next or if im doing it right to begin with?
what next? Figure out the potential at the original corner, and the potential at the new corner
V=kQ/distance
the distance from the furthermost corner changes
I'm unsure of how to figure out the potential at the different corners because if the first distance is kq/1.5, why isnt the 2nd one kq/1.5?
Im doing (9e9)(2.00e-6)/1.5=12000, which I don't know which number that is, or what to do with it further?
If your sides are equal, the original distance to the corner is found by the hypotenuese of the sides. Which is 2.1213m This is your original to the far corner, when moving to the side it comes down to 1.5. Plug those numbers in.
To find the work done by the electric force as one of the charges moves to an empty corner, you can use the formula for the work done by a conservative force:
W = -ΔPE
Where W is the work, ΔPE is the change in potential energy, and the negative sign indicates that work is done against the electric force.
To determine the change in potential energy, you can use the formula:
ΔPE = q * (Vf - Vi)
Where q is the charge and Vf and Vi are the final and initial potential energies respectively.
In this case, you have two identical charges (+2.0 µC) and one charge is being moved. The other two charges on the diagonal corners will remain fixed. So, we can consider the two fixed charges as creating a potential at the empty corner.
The potential at a point due to a point charge is given by the formula:
V = k * (q / r)
Where k is the electrostatic constant (8.99 x 10^9 Nm^2/C^2), q is the charge, and r is the distance from the charge.
In the initial configuration, the potential at the empty corner (Vi) is the sum of the potentials due to the two fixed charges. Since the charges are diagonal to each other, the distance between the charge and the empty corner is 1.5√2 m.
Vi = k * (q / r1 + q / r2)
Where r1 is the distance from the first fixed charge to the empty corner and r2 is the distance from the second fixed charge to the empty corner.
In the final configuration, after moving one of the charges to the empty corner, the potential at the empty corner (Vf) is now due to only the remaining fixed charge.
Vf = k * (q / r)
Where r is the distance from the remaining fixed charge to the empty corner.
Now, you can substitute the values into the equation for ΔPE:
ΔPE = q * (Vf - Vi)
= q * (k * (q / r) - k * (q / r1 + q / r2))
= q^2 * k * (1 / r - 1 / r1 - 1 / r2)
Substitute the known values: q = 2.0 µC (2.0 x 10^-6 C)
r1 = 1.5√2 m, r2 = 1.5√2 m, and r = 1.5 m
Now, calculate ΔPE using these values.
Finally, use the formula W = -ΔPE to find the work done by the electric force.