Question

An electric potential field is probed over a space and is found to have the form, $V(r) = -\frac{\lambda}{2\pi\epsilon_0}ln(\frac{r}{R})$ where $\lambda$ and R are constants A.) Derive the electric field vector $\vec{E}$ that is linked to this potential field in terms of the radial distance $r$ and constants B.) Is there a distance $r$ at which this result breaks down? Why? What geometry is creating this field?

          An electric potential field is probed over a space and is found to have the form,
$V(r) = -\frac{\lambda}{2\pi\epsilon_0}ln(\frac{r}{R})$
where $\lambda$ and R are constants
A.) Derive the electric field vector $\vec{E}$ that is linked to this potential field in terms of the radial distance $r$ and constants
B.) Is there a distance $r$ at which this result breaks down? Why? What geometry is creating this field?

An electric potential field is probed over a space and is found to have the form,
V(r) = -(λ)/(2πϵ0)ln((r)/(R))
where λ and R are constants
A.) Derive the electric field vector E⃗ that is linked to this potential field in terms of the radial distance r and constants
B.) Is there a distance r at which this result breaks down? Why? What geometry is creating this field?

Added by Leah B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Emily Anderson

Step 1

So, we need to take the gradient of the given potential field. The potential field is given as V = A/r + B/r^2, where A and B are constants. To take the gradient, we need to find the partial derivatives of V with respect to each coordinate (x, y, z). However, Show more…

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An electric potential field is probed over a space and is found to have the form, where and are constants. Derive the electric field vector that is linked to this potential field in terms of the radial distance r and constants. Is there a distance r at which this result breaks down? Why? What geometry is creating this field?