Equipotential Points from Electric Field
In the plane, there is a two-dimensional electric field acting over the entire plane:
On the line , there are two points at the same electric potential as the origin. What is the product of the -coordinates of those two points?
The answer is -3.0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
It is easy to verify that the curl of the vector field is zero. Therefore, the vector field can be written as the gradient of a potential function.
∂ x ∂ U = E x = 2 x ∂ y ∂ U = E y = − 3
By inspection, the first condition yields:
U = x 2 + g ( y ) + C
The second equation yields:
g ( y ) = − 3 y + D ⟹ U = x 2 − 3 y + D + C U = x 2 − 3 y + E
The potential at the origin is clearly E . For another point to equal the origin's potential, the following condition must hold:
x 2 − 3 y = 0
Along the line ( y = 1 − x ) , this can be written as:
x 2 − 3 ( 1 − x ) = 0 x 2 + 3 x − 3 = 0 x = 2 − 3 ± 2 1
The product of the two solutions is − 3