The figure shows two equipotential lines in t | vedclass

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

Question

(B) The equation of the equipotential lines is $V = ax + by + c$. From the graph,for $V = 2 \, V$,the line passes through $(6, 2)$ and $(4, 1)$. The s

Vedclass · Accepted Answer

$-100 \, V/m, +200 \, V/m$

Vedclass · Answer

(B) The equation of the equipotential lines is $V = ax + by + c$. From the graph,for $V = 2 \, V$,the line passes through $(6, 2)$ and $(4, 1)$. The slope is $m = \frac{1-2}{4-6} = \frac{-1}{-2} = 0.5$. The equation is $y - 1 = 0.5(x - 4) \Rightarrow y = 0.5x - 1 \Rightarrow 0.5x - y = 1$. Multiplying by $2$,we get $x - 2y = 2$. Thus,$V = k(x - 2y)$. For $V = 2$,$2 = k(6 - 2(2)) = 2k \Rightarrow k = 1$. So,$V = x - 2y$. The electric field components are $E_x = -\frac{\partial V}{\partial x} = -1 \, V/	ext{cm} = -100 \, V/m$ and $E_y = -\frac{\partial V}{\partial y} = -(-2) = +2 \, V/	ext{cm} = +200 \, V/m$. Thus,$E_x = -100 \, V/m$ and $E_y = +200 \, V/m$.

The figure shows two equipotential lines in the $x, y$ plane for an electric field. The scales are marked in $\text{cm}$. The $x$-component $E_x$ and $y$-component $E_y$ of the electric field in the space between these equipotential lines are respectively:

Similar Questions

What is potential gradient?

If the electric potential at any point $(x, y, z) \, m$ in space is given by $V = 3x^2$ volt,the electric field at the point $(1, 0, 3) \, m$ will be ............

Electric potential at any point is $V = -5x + 3y + \sqrt{15} z$,then the magnitude of the electric field is

The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (4x^2 + 8x - 3) \ V$. The value of the electric field at $x = 0.5 \ m$,in $V/m$,is:

The electric potential at a point in free space due to charge $Q$ coulomb is $V = Q \times 10^{11} \, V$. The electric field at that point is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module