The electric potential Vin the space between two flat parallel plates 1 and 2 is given (in volts) by V = 1500x², where x(in meters) is the perpendicular distance from plate 1. At x = 1.3cm, (a) what is the magnitude of the electric field and (b) is the field directed toward or away from plate 1?

Using the given data of the electric potential and the given distance in the equation of the electric field, we can get the magnitude and the directional value of the electric field at the given point.

Formula:

The relation between electric field and electric potential is given by, $E=-\frac{\delta V}{\delta x}$ (i)

Using the given data, we can get the magnitude of the electric field at x = 1.3cm as follows:

$\begin{array}{rcl}E& =& -\frac{\delta \left(1500x^2\right)}{\delta x}\\ & =& -3000x\\ & =& -3000\left(1.3cm\right)\\ & =& -3000\left(1.3cm\right)\left(\frac{{10}^{-2}m}{1cm}\right)\\ & =& -39N/C\end{array}$

Therefore, the magnitude of the electric field is 39N/C.

The electric field is equal to the negative rate at which the potential changes with distance.

$E=-\frac{\delta V}{\delta x}$

Negative sign indicates that the direction of an electric field is from high potential to low potential. The potential increases with an increase in distance from the plate 1 (because V = 1500x²). So the potential increases from plate 1 to plate 2 and hence the direction of the electric field is from plate 2 to plate 1.

Therefore, the direction of the electric field is towards plate 1.