Two point charges, +3.18 µC and -6.40 µC, are separated by 1.32 m. What is the electric potential midway between them?
v=kq/r
r=.5(1.32)=0.66m
k=(8.99*10^9)
q1= 3.18x10^-6 C
q2= -6.40x10^-6 C
v1= (k(3.18x10^-6C)/0.66m)=4.3x10^4v
v2= (k(-6.4x10^-6C)/0.66m)=-8.8x10^4v
v1+v2=v(total)= -4.5x10^4 v
To find the electric potential midway between two point charges, we can use the formula:
V = k * (q1 / r1 + q2 / r2)
where:
V is the electric potential,
k is the Coulomb's constant (9.0 x 10^9 N m^2/C^2),
q1 and q2 are the magnitudes of the charges, and
r1 and r2 are the distances from the charges to the midpoint.
In this case, q1 = +3.18 µC, q2 = -6.40 µC, and the distance between the charges is 1.32 m. To find the electric potential at the midpoint, we need to determine the distances from the charges to the midpoint.
Since the charges are equally spaced, the distance from each charge to the midpoint is half of the distance between them, which is 1.32 m / 2 = 0.66 m.
Now we can substitute the values into the formula:
V = (9.0 x 10^9 N m^2/C^2) * (3.18 µC / 0.66 m + (-6.40 µC) / 0.66 m)
Calculating this expression will give us the electric potential at the midpoint.