The question states: two large, parallel conducting plates are 12cm
apart and have charges of equal magnitude and opposite sign ontheir
facing surfaces. An electrostatic force of 3.9 x 10^-15 Nacts on an
electron placed anywhere between two plates (neglectingthe
fringing).
1. find the elctric field at the position of the elctron.
2. what is the potential difference between the plates?

Question

contestada

Respuesta :

Vespertilio Vespertilio · Answer 1 · 2019-09-20T12:40:47+02:00

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, [tex]E=\frac{F}{q}[/tex]

[tex]E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}[/tex]

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt