Two point charges are placed on the x axis. The charge +4q is at x=3.00 m, and the -q charge
is at x=-3.00.
a) There is a point on the x axis between the two charges where the electric potential is zero. Where is this point?
Found it to be: -1.80m
b) The electric potential also vanishes in one of the following regions:
Region 1, 10.0m>x>3.00 m;
Region 2, -3.00m>x> -10.0.
Identify the appropriate region.
Found to be region 2.
c) Find the value of x referred to in part B.
Not sure how to do this part.
V=kq1/r1 + kq2/r2=k(+4/(x+3)-1/x-3) where x is between-3 and -10
solving
4(x-3)=1(x+3)
3x=15
x=5m, which in coordinate terms is x=-5.00
To find the value of x referred to in part B, we need to find the boundary between Region 2 and Region 1 where the electric potential vanishes.
The electric potential due to a point charge q at a distance r from the charge is given by the equation V = k*q/r, where k is the electrostatic constant (approximately 9 x 10^9 Nm^2/C^2).
In Region 1, the electric potential is zero when V = k*q1/r1 + k*q2/r2 = 0, where q1 = +4q is the charge at x = 3.00 m and q2 = -q is the charge at x = -3.00 m.
In Region 2, the electric potential is zero when V = k*q2/r2 = 0, where q2 = -q is the charge at x = -3.00 m.
To find the value of x, we need to set up equations using these conditions.
In Region 1:
k*q1/r1 + k*q2/(10.0m - x) = 0
In Region 2:
k*q2/r2 = 0
Let's solve these equations to find the value of x.
From the equation in Region 1:
k*q1/r1 = - k*q2/(10.0m - x)
Substituting the known values:
k*(+4q)/(3.00m) = - k*(-q)/(10.0m - x)
Simplifying:
4/(3.00m) = - 1/(10.0m - x)
Cross-multiplying:
-4(10.0m - x) = 3.00m
Expanding the equation:
-40.0m + 4x = 3.00m
Combining like terms:
4x = 3.00m + 40.0m
Simplifying:
4x = 43.00m
Dividing both sides by 4:
x = 43.00m / 4
Calculating:
x ≈ 10.75m
So, the value of x referred to in part B is approximately 10.75m, which falls within Region 2 (-3.00m > x > -10.0m).