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

Answer on Question #86939, Physics / Electric Circuits

Given:

$$\mathrm{D} = 0.75 \mathrm{m}. \mathrm{F} = 4.8 \times 10^{-16} \mathrm{N}.$$

To find:

(a) The magnitude of the electric field at the position of the electron between the plates.

(b) The potential difference between the plates.

Solution:

(a) Magnitude of electric field at the position of electron between the plates, we use $E = \frac{F}{q}$ where $q$ is an elementary charge carried by single electron and $F$ is the force. The elementary charge is $q = 1.6 * 10^{-19} C$.

Thus,

$$E = \frac{4.8 \times 10^{-16}}{1.6 \times 10^{-19}}$$ $$E = 3000 \mathrm{N/C}$$

(b) The potential difference between the plates, we multiply the electric field value by the distance between the plates.

$$V = E \cdot D$$ $$V = (3000 \mathrm{N/C}) \cdot (0.75 \mathrm{m})$$ $$V = 2250 \mathrm{V}$$

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