Question

Calculate the electric field (in V/m) between two parallel plates connected to a 45 V battery if the plates are separated by a) 3 cm and b) 5cm. What is the resulting force on an electron in each of these fields?

Answer 1

Answer:

Explanation:

Given

Voltage of battery $V=45\ V$

electric field(E) is given by  

$E=|-\frac{dV}{dx}|$

i.e. Change in Potential over a distance x

(a)When Plate are separated by 3 cm apart

$E=|\frac{45}{3\times 10^{-2}}|$

$E=1500\ V/m$

(b)When Plates are 5 cm apart

$E=|-\frac{45}{5\times 10^{-2}}|$

$E=900\ V/m$

Force on electron is given by

$F=charge\times Electric\ Field$

$F=q\times E$

For case a

$F=1.6\times 10^{-19}\times 1500$

$F=24\times 10^{-17}\ N$

(b)$F=1.6\times 10^{-19}\times 900=14.4\times 10^{-17}\ N$

Answer 2

Answer:

Explanation:

Voltage, V = 45 V

The relation between the electric field and the distance is

E = V / r

(a) r = 3 cm = 0.03 m

E = 45 / 0.03

E = 1500 V/m

(b) r = 5 cm = 0.05 m

E = 45 / 0.05

E = 900 V/m