Two infinitely-long lines of charge run parallel to the z axis. One has positive unifonn charge per unit length, IvO, and goes through the x y plane at x=O, y=s/2. The other has negative unifonn charge per unit length, -A, and goes through x=O, y=-s/2. Nothing changes with the z coordinate; the state of affairs in any plane parallel to the x y plane is the same in the x y plane.

a. Describe the vector, R+, going from the positive line ofcharge to a
generic point, (x,y), and the vector R_ from the negative line.
b.
Find the total electric field at the point (x,y) as a vector.
c.
Find the voltage, V=V(x,y).
d.
To begin looking for equipotentials, set the voltage equal to a constant
called V = ~ In(a). Simplify and exponentiate to remove InO.
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e.
Multiply through by any denominators, then move everything to the left side and simplify.
f.
Find the multiplier, "2yc' " that multiplies y, and "complete the square"
by adding 2ycY to both sides.
g. Write the equation of an equipotential in the fonn x2 + (y -y c)2 = R2 and
find the radius and center.
h. Find numerical values for the voltage, center and radius for a=30, 10,3, I, 1/3, 1110, and 1/30.
1. Using s=four squares, draw these circles on graph paper.
J. Draw four or five lines of force of the electric field.

a. To describe the vectors R+ and R_, we can consider the coordinates of a generic point (x, y).

For the positive line of charge, the vector R+ is the position vector going from the positive line of charge to the point (x, y). Since the positive line of charge is located at (x=0, y=s/2), the vector R+ can be expressed as R+ = xi + (y - s/2)j, where i and j are unit vectors in the x and y directions, respectively.

Similarly, for the negative line of charge, the vector R_ is the position vector going from the negative line of charge to the point (x, y). Since the negative line of charge is located at (x=0, y=-s/2), the vector R_ can be expressed as R_ = xi + (y + s/2)j.

b. To find the total electric field at the point (x, y), we need to consider the contributions from both the positive and negative lines of charge. The electric field due to a line of charge is given by the expression:

E = (λ / (2πε₀)) * (1 / R),

where λ is the charge per unit length, ε₀ is the permittivity of free space, and R is the distance between the point and the line of charge.

Given that λ₀ is the positive charge per unit length and -A is the negative charge per unit length, the total electric field at the point (x, y) can be found by adding the contributions from both lines:

E_total = E_+ - E_,

where E_+ = (λ₀ / (2πε₀)) * (1 / |R+|) and E_ = (A / (2πε₀)) * (1 / |R_|).

c. To find the voltage, V(x, y), we can integrate the electric field E_total with respect to the position vector R:

V(x, y) = -∫E_total · dR,

where dR is an infinitesimal displacement along the position vector R.

d. Setting the voltage equal to a constant V = ~ In(a), we can rearrange the equation V(x, y) = ~ In(a) and solve for a:

a = exp(V / ~ ).

e. To remove any denominators, we can multiply both sides of the equation by 2πε₀:

2πε₀ * V(x, y) = 2πε₀ * ~ In(a).

f. Moving everything to the left side and simplifying, we get:

2πε₀ * V(x, y) - 2πε₀ * ~ In(a) = 0.

Then, we can find the multiplier "2yc" by comparing the coefficient of y in the equation above to the expression 2ycY, and complete the square by adding 2ycY to both sides.

g. The equation of an equipotential in the form x² + (y - yc)² = R² represents a circle with center (0, yc) and radius R.

To find the radius and center, we can compare the equation obtained in step f to the equation of the circle and extract the values of R and yc.

h. By substituting the given values of a=30, 10, 3, 1, 1/3, 1110, and 1/30 into the equation obtained in step g, we can find numerical values for the voltage, center, and radius.

1. To draw the equipotential circles, use graph paper and plot the circles with the obtained values for center (0, yc) and radius R.

J. To draw the lines of force of the electric field, consider the definition of the electric field as the negative gradient of the voltage:

E = -∇V(x, y),

where ∇ is the gradient operator. Draw the lines of force, which are tangential to the electric field vectors, on the graph paper to represent the direction and strength of the electric field at different points.