Question

The electric potential V in the space between two flat parallel plates 1 and 2 is given (in volts) by V = 2200x2, where x (in meters) is the perpendicular distance from plate 1. At x = 1.7 cm, (a) what is the magnitude of the electric field and (b) is the field directed toward or away from plate 1?

Answer 1

Answer:

a) 74.8 V/m

b) direction -(i)

Explanation:

For this case, the electric field is defined as:

$E=(\frac{dV}{dx})$

because V is a function dependent of x so we can derivate it:

$\frac{dV}{dx}=\frac{d(2200x^2)}{dx}=2*2200*x$

so:

$E=-4400x\\E=-4400V/m^2*0.017m=-74.8V/m$

Because of the sign, the magnitude of electric field 74.8 V/m on the negative direction of i vector, thus means towards the plate 1.