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.
4n
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. The vector R+ going from the positive line of charge to a generic point (x, y) can be determined by considering the coordinates of the point relative to the positive line. Since the positive line runs through the x-y plane at x=0, y=s/2, the vector R+ can be written as R+ = x i + (y - s/2) j, where i and j are the unit vectors in the x and y directions, respectively.
Similarly, the vector R- going from the negative line of charge to the same generic point (x, y) is given by R- = x i + (y + s/2) j, since the negative line runs through the x=0, y=-s/2.
b. To find the total electric field at the point (x, y), we need to consider the contribution from both the positive and negative lines of charge.
The electric field due to an infinitely long line of charge with charge density λ at a distance r from the line is given by E = λ / (2πε₀r), where ε₀ is the permittivity of free space.
Using this formula, the electric field E+ due to the positive line of charge at point (x, y) is given by E+ = (λ₀ / (2πε₀)) * (x i + (y - s/2) j) / ((x² + (y - s/2)²)^(3/2)), where λ₀ is the charge per unit length of the positive line.
Similarly, the electric field E- due to the negative line of charge at point (x, y) is given by E- = (-λ₀ / (2πε₀)) * (x i + (y + s/2) j) / ((x² + (y + s/2)²)^(3/2)).
The total electric field can then be obtained by adding these two electric field vectors: E_total = E+ + E-.
c. To find the voltage V at the point (x, y), we can integrate the electric field along a path from a reference point to (x, y). The voltage at a point is defined as the work done per unit charge to bring a positive test charge from a reference point to that point in an electric field.
Thus, V(x, y) = - ∫E_total · dl, where dl is the differential displacement along the path of integration.
d. To begin looking for equipotentials, we set the voltage equal to a constant V = ln(a).
e. We can now simplify and exponentiate to remove the natural logarithm: a = exp(V).
f. We multiply through by any denominators and move everything to the left side to simplify the equation obtained in the previous step.
g. To "complete the square" and write the equation in the form x^2 + (y - yc)^2 = R^2, we need to find the multiplier that multiplies y and add 2ycy to both sides of the equation from the previous step.
h. Once we have the equation of an equipotential in the form x^2 + (y - yc)^2 = R^2, we can identify the radius (R) and center (xc, yc) of the equipotential.
i. Using the numerical values provided for a, we can substitute them into the equation of the equipotential obtained in the previous steps to find the specific values for voltage, center, and radius.
j. To draw the circles on graph paper using four squares, use the equation of the equipotential x^2 + (y - yc)^2 = R^2 and substitute the values obtained in step h.
k. To draw lines of force of the electric field, we can use the electric field vectors obtained in step b. Plot the vectors at various points in the coordinate plane (x, y) and draw lines that are tangential to these vectors. Repeat this process for different points to get multiple lines of force.