The electric field in the xy-plane due to an infinite line...

The electric field in the xy-plane due to an infinite line of charge along the z-axis is a gradient field with a 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 $x y$ -plane is directed outward from the origin and has magnitude $|\mathbf{E}|=c / r,$ 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 $V$

The electric field in the xy-plane due to an infinite line of charge along the z-axis is a gradient field with a 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 $x y$ -plane is directed outward from the origin and has magnitude $|\mathbf{E}|=c / r,$ 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 $V$

Key Concepts

Gradient and Potential Functions

This concept involves understanding that the electric field is the negative gradient of the electric potential, meaning that E is derived by differentiating the potential function with respect to spatial coordinates. It explains how variations in the potential give rise to the electric field at each point in space.

Infinite Line of Charge

This refers to a charge distribution that extends indefinitely in one dimension, producing a symmetrical electric field in the perpendicular plane. The treatment of such a distribution uses cylindrical symmetry, leading to a logarithmic potential function in the plane perpendicular to the line charge.

Electric Field Magnitude and Direction

This concept highlights that the magnitude of the electric field can be determined by the rate of change (gradient) of the potential, and its direction is given by the sign and vector of that rate of change. In symmetric charge distributions like an infinite line, the field magnitude often takes a form inversely proportional to the distance from the charge.

Orthogonality of Electric Field and Equipotential Curves

Equipotential curves are loci of points with the same potential. The electric field, being perpendicular to these curves, demonstrates that no work is done when moving along an equipotential, which corroborates the relationship between potential and field. This property is critical in visualizing and solving problems involving fields and potentials in electrostatics.

Transcript

00:01 In this problem of vector calculus we have given that the electric field in the xy plane so xy plane due to an infinite light of line of charge along the z axis is gradient field with a potential function say v which is the function of x and y is equals to c ln multiplied with r not divided with under root of x square plus o y is square where c is greater than zero and here we have given that c is greater than zero is a constant and r not is a reference distance at which the potential is assumed to be zero so for this we have given a figure let me show you so this is the figure and we have to find the component of the electric field in x and y direction where we have given that e x y is equal to negative of grad v so first we have to differentiate this term so this would be say e is equals to now when we differentiate a differentiation of this term is c is treated as constant and this would be here x divided with this would be x squared plus y square and when we differentiate this term with respect to y so this is y divided with x square plus y is square so this is the required term or this would be here this term is multiplied with i cap plus and this would be multiplied with j cap and we can write vector e is equals to this is c divided with this is x square plus y square and this is x and y so this is the required vector field and the solution for part a now part b so part b says here we have to show that the electric field at a point in x y plane is directed outwards from the origin and has the magnitude say magnitude is equals to c divided with r where r is equal to under root of x square plus y square so electric field e is equals to this is modulus of c divided with modulus of r squared multiplied with r so from here one r can be cancelled out and modulus of e is equals to c divided with r so this is we have proved so this is the solution for part b and this is the solution for part b.

03:16 And now, and we have to show that the vector field is orthogonal to equipotential at all points in the domain of v...

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