The variation of potential V with distance x | vedclass

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

Question

(C) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$,which is the negative of the slope of the $V-x$ graph.

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dx}$,which is the negative of the slope of the $V-x$ graph.At $x = 13\,m$,the point lies on the line segment between $x = 12\,m$ and $x = 16\,m$.The coordinates of the endpoints of this segment are $(12, 30)$ and $(16, 10)$.The slope of this line segment is $	ext{slope} = \frac{V_2 - V_1}{x_2 - x_1} = \frac{10 - 30}{16 - 12} = \frac{-20}{4} = -5\,V/m$.Therefore,the electric field $E = -(	ext{slope}) = -(-5) = 5\,V/m$.

The variation of potential $V$ with distance $x$ from a fixed point is as shown in the figure. The electric field at $x = 13\,m$ is......$V/m$.

Similar Questions

If the electric potential is given by $V = 4x + 3y$,what is the electric field at the point $(2, 1)$?

The variation of electric potential $V$ with distance $x$ from a fixed point is shown in the figure. What is the value of the electric field at $x = 2 \ m$?

If the electric potential is given by $V = (5x^2 + 10x - 9) \ V$,what is the electric field at $x = 1 \ m$ in $V/m$?

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?

The electric potential inside a charged sphere is given by $\phi = ar^2 + b$,where $r$ is the distance from the center,and $a$ and $b$ are constants. The charge density inside the sphere is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module