The three charges in the figure are at the vertices of an isosceles triangle. Calculate the work that needs to be expanded, in Joules, to bring the charges together (as shown.) Take q=5.7 μC. The length of base is b=2.30 m. The lengths of sides are s=5.80 m. The two charges at the base are negative (q1=q2=-q), the third charge is positive (q3=+q)
To calculate the work required to bring the charges together, we need to determine the potential energy of the system. The potential energy of a system of charges is given by the equation:
U = (1/4πε₀) * (q1q2/r1 + q1q3/r2 + q2q3/r3)
where U is the potential energy, q1, q2, q3 are the charges, r1, r2, r3 are the distances between the charges, and ε₀ is the permittivity of free space.
In this case, we have q1 = q2 = -q and q3 = +q, where q is given as 5.7 μC (microCoulombs). We also have the distance between the charges on the base, which is b = 2.30 m, and the distance between each side charge and the top charge, which is s = 5.80 m.
First, we can calculate the distances between the charges. Since the triangle is isosceles, the two base charges are equidistant from the top charge. Therefore, their distance from the top charge is half the length of the base, which is b/2.
r1 = r2 = b/2 = 2.30 m / 2 = 1.15 m
The distance between the side charges and the top charge is given as s.
r3 = s = 5.80 m
Now we can substitute the values into the equation for potential energy:
U = (1/4πε₀) * (-q²/1.15 + (-q)(q)/1.15 + (-q)(q)/5.80)
To calculate the work done to bring the charges together, we can use the relationship between work and potential energy:
W = -ΔU
Since we are bringing the charges together, the potential energy decreases, so we add a negative sign.
W = -(-ΔU) = ΔU
Therefore, the work required to bring the charges together is equal to the change in potential energy of the system.
Now, we can calculate the work required by substituting the given values into the equation for potential energy.
W = (1/4πε₀) * (-q²/1.15 + (-q)(q)/1.15 + (-q)(q)/5.80)
Let's calculate the value of W using these values:
ε₀ = 8.854 x 10^-12 C²/N·m² (permittivity of free space)
q = 5.7 x 10^-6 C (charge)
Substituting these values and evaluating the expression will give us the work required in Joules.