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Electric field due to a line of charge The electric field
Chapter 14, Problem 45E(choose chapter or problem)
Electric field due to a line of charge The electric field in the xy-plane due to an infinite line of charge along the z-axis is a gradient field with u potential function \(V(x, y)=c \ln \left(\frac{r_{0}}{\sqrt{x^{2}+y^{2}}}\right)\), where c > 0 is a constant and \(r_{0}\) is a reference distance at which the potential is assumed to be 0 (see figure).
a. Find the components of the electric field in the x- and y- directions, where \(\mathbf{E}(x, y)=-\nabla V(x, y)\).
b. Show that the electric field at a point in the xy-plane is directed outward from the origin and has magnitude \(|\mathbf{E}|=c / r \text {, where } r=\sqrt{x^{2}+y^{2}}\).
c. Show that the vector field is orthogonal to the equipotential curves at all points in the domain of V.
Questions & Answers
QUESTION:
Electric field due to a line of charge The electric field in the xy-plane due to an infinite line of charge along the z-axis is a gradient field with u potential function \(V(x, y)=c \ln \left(\frac{r_{0}}{\sqrt{x^{2}+y^{2}}}\right)\), where c > 0 is a constant and \(r_{0}\) is a reference distance at which the potential is assumed to be 0 (see figure).
a. Find the components of the electric field in the x- and y- directions, where \(\mathbf{E}(x, y)=-\nabla V(x, y)\).
b. Show that the electric field at a point in the xy-plane is directed outward from the origin and has magnitude \(|\mathbf{E}|=c / r \text {, where } r=\sqrt{x^{2}+y^{2}}\).
c. Show that the vector field is orthogonal to the equipotential curves at all points in the domain of V.
ANSWER:
Solution 45E
Step 1
