The electric potential V(x, y, z) for a plana | vedclass

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

Question

(A) The electric field $E_x$ is related to the potential $V$ by the relation $E_x = -\frac{dV}{dx}$.$1$. For $x < -d$,$V = 0$,so $E_x = -\frac{d(0)}{d

Vedclass · Accepted Answer

Vedclass · Answer

(A) The electric field $E_x$ is related to the potential $V$ by the relation $E_x = -\frac{dV}{dx}$.$1$. For $x $2$. For $-d \le x $E_x = -\frac{d}{dx}[-V_0(1 + \frac{x}{d})^2] = V_0 \cdot 2(1 + \frac{x}{d}) \cdot \frac{1}{d} = \frac{2V_0}{d}(1 + \frac{x}{d})$.At $x = -d$,$E_x = 0$. At $x = 0$,$E_x = \frac{2V_0}{d}$.$3$. For $0 \le x $E_x = -\frac{d}{dx}[-V_0(1 + 2\frac{x}{d})] = V_0 \cdot \frac{2}{d} = \frac{2V_0}{d}$.This is a constant value.$4$. For $x \ge d$,$V = -3V_0$,so $E_x = -\frac{d(-3V_0)}{dx} = 0$.Comparing these results with the given options,the graph in option $A$ correctly represents $E_x$ as a function of $x$.

Similar Questions

The figure shows the electric potential $V$ as a function of distance through five regions on the $x$-axis. Which of the following is true for the electric field $E$ in these regions?

Assume that an electric field $E=20 x^2 \hat{i}$ exists in space. If $V_0$ is the potential at the origin and $V_A$ is the potential at $x=3 \ m$,then the potential difference $V_A-V_0$ in volts is

For the given figure,the value of the electric field is ......... $V/m$.

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module