Figure shows two equipotential lines in the xy plane for an electrical field. The scales are marked. The x-component and y-component of the field in the space between these equipotential lines are respectively :

Question

Verified by Toppr

Answer 1

The relation between Electric field and potential is $E = - \nabla V = - \frac{\partial V}{\partial x} \hat{i} - \frac{\partial V}{\partial y} \hat{j}$

From the given graph, $V (4, 1) = 2 V$ and $V (6, 1) = 6 V$
Thus, $\frac{\partial V}{\partial x} = \frac{V ( 6 , 1 ) - V ( 4 , 1 )}{6 - 4} = \frac{6 - 2}{2} = 2 V/cm = 200 V/m$
Thus, $E_{x} = - \frac{\partial V}{\partial x} = - 200 V/m$

From the given graph, $V (6, 2) = 2 V$ and $V (6, 1) = 6 V$
Thus, $\frac{\partial V}{\partial y} = \frac{V ( 6 , 2 ) - V ( 6 , 1 )}{2 - 1} = - 4 V/cm = - 400 V/m$
$∴$ $E_{y} = - \frac{\partial V}{\partial y} = 400 V/m$

Answer 2

The relation between Electric field and potential is

E = - \nabla V = - \frac{\partial V}{\partial x} \hat{i} - \frac{\partial V}{\partial y} \hat{j}

From the given graph,

V (4, 1) = 2 V

and

V (6, 1) = 6 V

Thus,

\frac{\partial V}{\partial x} = \frac{V ( 6 , 1 ) - V ( 4 , 1 )}{6 - 4} = \frac{6 - 2}{2} = 2 V/cm = 200 V/m

Thus,

E_{x} = - \frac{\partial V}{\partial x} = - 200 V/m

From the given graph,

V (6, 2) = 2 V

and

V (6, 1) = 6 V

Thus,

\frac{\partial V}{\partial y} = \frac{V ( 6 , 2 ) - V ( 6 , 1 )}{2 - 1} = - 4 V/cm = - 400 V/m

∴

E_{y} = - \frac{\partial V}{\partial y} = 400 V/m