The electric potential V at any point (x, y, | vedclass

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

Question

(A) The electric potential is given by $V(x, y, z) = 4x^2 	ext{ volt}$.The electric field $\vec{E}$ is related to the potential $V$ by the relation $

Vedclass · Accepted Answer

$8$ along negative $X$-axis

Vedclass · Answer

(A) The electric potential is given by $V(x, y, z) = 4x^2 	ext{ volt}$.The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -
abla V = -\left( \hat{i} \frac{\partial V}{\partial x} + \hat{j} \frac{\partial V}{\partial y} + \hat{k} \frac{\partial V}{\partial z} ight)$.Calculating the partial derivatives:$\frac{\partial V}{\partial x} = \frac{d}{dx}(4x^2) = 8x$$\frac{\partial V}{\partial y} = 0$$\frac{\partial V}{\partial z} = 0$Substituting these into the expression for $\vec{E}$:$\vec{E} = -(8x \hat{i} + 0 \hat{j} + 0 \hat{k}) = -8x \hat{i} 	ext{ V/m}$.At the point $(1 	ext{ m}, 0, 2 	ext{ m})$,the $x$-coordinate is $1 	ext{ m}$.Therefore,$\vec{E} = -8(1) \hat{i} = -8 \hat{i} 	ext{ V/m}$.The negative sign indicates that the electric field is directed along the negative $X$-axis with a magnitude of $8 	ext{ V/m}$.

The electric potential $V$ at any point $(x, y, z)$ (all in metres) in space is given by $V = 4x^2 \text{ volt}$. The electric field at the point $(1 \text{ m}, 0, 2 \text{ m})$ in $\text{volt/metre}$ is

Similar Questions

The figure shows the variation of electric field intensity $E$ versus distance $x$. What is the potential difference between the points at $x = 2 \, m$ and $x = 6 \, m$ from $O$ (in $V$)?

If $V$ is the electric potential at a given point,then the electric field $E_x$ in the $x$-direction at that point is:

The electrostatic potential inside a charged sphere is given as $V = A r^2 + B$,where $r$ is the distance from the centre of the sphere,$A$ and $B$ are constants. Then,the charge density in the sphere is

Let $V$ and $E$ be the potential and electric field intensity at a point,respectively. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module