Question

diagram above each unit on the horizontal axis is 3.00 cm and each unit on the vertical axis is 3.00 cm. The equipotential lines in a region A) Determine the magnitude of the electric field in this region. V/m B)Determine the shortest distance for which the change in potential is 3 V.

Answer 1

Electric field in x-direction can be written as:

$E_{x}=-\frac{dV}{dx}$   

We call the quantity $\frac{dV}{dx}$, the gradient of the electric potential in the -direction. It basically measures how fast the potential varies as the coordinate is changed (but the coordinates y and are held constant).

Similarly: $E_{y}=-\frac{dV}{dy}$ ,   $E_{z}=-\frac{dV}{dz}$

From the graph:

$E_{x}=-\frac{dV}{dx} =-\frac{12-6}{6-3}$

$E_{x}=-\frac{6}{3}$

This is in x-direction, Now find in y-direction as well

$E_{y}=-\frac{dV}{dy} =-\frac{12-6}{-6-(-3)}$

$E_{y}=-\frac{6}{-3} = \frac{6}{3}$

Therefore

magnitude of E = 282.8 V/m ....... Ans to partA

ANSWER to PARTB:

$E=-\frac{dV}{dr}$

change in potential = dV = 3

E = 282.8V/m

therefore:

$dr=\frac{3}{282.8}$ m

Here, we can drop the -ve sign bcz it just indicates that potential decreases in the direction of electric field

dr = 0.01m = 1cm