Answer to Question #86764 in Electricity and Magnetism for sahib

Question

Two large parallel conducting plates carrying equal and opposite charges on their
surfaces facing each other are placed at a distance of 0.75 m. An electron placed at
some point between the plates experiences an electrostatic force of 4.8 × 10^-16 N.
Calculate the magnitude of the electric field at the position of the electron between
the plates. Also determine the potential difference between the plates.

Answer 1

Find: E – ? V – ?

Given:

d=0.75 m

F=4.8 × 10^-16 N

q=1.6× 10^-19 C

Solution:

The magnitude E of the electric field:

$$E=F\div q (1)$$

where F is the electric force, q is the electric charge

Using the formula (1) we get: E=3000 N/С

The potential difference between the plates:

$$V=E\times d (2)$$

Using the formula (2) we get: V=2250 V

Answer:

the magnitude of the electric field: 3000 N/С

the potential difference between the plates:2250 V