$V_{0}$ is the potential at the origin in an electric field. $E = E_{x} i - E_{y} j$ . The potential at the point (x,y) is:

Question

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Answer 1

We know
$V (x_{1}) - V (x_{2}) = \int_{x_{1}}^{x_{2}} E d x .$
Now potential change in x direction
$V_{x} (x) - V_{x} (o) = \int_{x}^{o} E_{x} d x = - E_{x} x .$
$V_{x} (x) = V_{x} (o) - E_{x} x$
Potential change in y direction
$V_{y} (y) - V_{y} (o) = \int_{y}^{o} (- E_{y}) d y = + E_{y} y$
$V_{y} (y) = V_{y} (o) + E_{y} y$
Now total potential at (x,y)
$= V_{x} + V_{y}$
$= V_{x} (o) - E_{x} x + V_{y} (o) + E_{y} y$
$= V_{o} - x E_{x} + y E_{y}$

Answer 2

We know

V (x_{1}) - V (x_{2}) = \int_{x_{1}}^{x_{2}} E d x .

Now potential change in x direction

V_{x} (x) - V_{x} (o) = \int_{x}^{o} E_{x} d x = - E_{x} x .

V_{x} (x) = V_{x} (o) - E_{x} x

Potential change in y direction

V_{y} (y) - V_{y} (o) = \int_{y}^{o} (- E_{y}) d y = + E_{y} y

V_{y} (y) = V_{y} (o) + E_{y} y

Now total potential at (x,y)

= V_{x} + V_{y}

= V_{x} (o) - E_{x} x + V_{y} (o) + E_{y} y

= V_{o} - x E_{x} + y E_{y}