The variation of potential with distance R fr | vedclass

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

Question

(A) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dR}$, which represents the negative of the slope of the $V-

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dR}$, which represents the negative of the slope of the $V-R$ graph.For the region between $R = 4\,m$ and $R = 6\,m$, the graph is a straight line passing through $(4, 5)$ and $(6, 0)$.The slope of this line is $m = \frac{V_2 - V_1}{R_2 - R_1} = \frac{0 - 5}{6 - 4} = \frac{-5}{2} = -2.5\,V/m$.Therefore, the electric field at $R = 5\,m$ is $E = -(	ext{slope}) = -(-2.5) = 2.5\,V/m$.

The variation of potential with distance $R$ from a fixed point is as shown in the graph. The electric field at $R = 5\,m$ is......$V/m$.

Similar Questions

If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$,the electric field (in $N/C$) at point $(1, 1, 0)$ is:

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

Due to some charges on the $x$-axis,the potential at a point on the $x$-axis is given by $V(x) = 20/(x^2 - 4) \ V$. The electric field at $x = 4 \ \mu m$ is given by:

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......$V/m$.

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module