The electric potential at a point (x, y) in t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The electric field components are given by the negative gradient of the potential:$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partia

Vedclass · Accepted Answer

$r$

Vedclass · Answer

(B) The electric field components are given by the negative gradient of the potential:$E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(-kxy) = ky$$E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(-kxy) = kx$The magnitude of the electric field intensity $E$ is given by:$E = \sqrt{E_x^2 + E_y^2} = \sqrt{(ky)^2 + (kx)^2}$$E = k\sqrt{x^2 + y^2}$Since the distance from the origin is $r = \sqrt{x^2 + y^2}$,we have:$E = kr$Therefore,the field intensity varies directly with the distance $r$ from the origin,i.e.,$E \propto r$.

The electric potential at a point $(x, y)$ in the $x-y$ plane is given by $V = -kxy$. The field intensity at a distance $r$ from the origin varies as

Similar Questions

Two large plane parallel conducting plates are kept $10 \ cm$ apart as shown in the figure. The potential difference between them is $V$. The potential difference between the points $A$ and $B$ (shown in the figure) is

The electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2$. Find the electric force in $N$ acting on a point charge of $2 \ C$ placed at the origin.

The potential $V$ varies with $x$ and $y$ as $V = \frac{1}{2}(y^2 - 4x) \text{ volts}$. The electric field at $(1 \text{ m}, 1 \text{ m})$ is:

The potential $V$ is varying with $x$ and $y$ as $V = \frac{1}{2}(y^2 - 4x) \text{ V}$. The electric field at $(1 \text{ m}, 1 \text{ m})$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module