The electric potential at 2 cm of a plate is V=12 V . If the plate measures 16 cm wide and 200 mm long. What is its electrical charge?

To find the electric charge on the plate, we first need to calculate the electric field at the surface of the plate.

Given:
- Electric potential, V = 12 V
- Distance from the plate, r = 2 cm = 0.02 m

We can use the formula relating electric potential to electric field:
V = E \times r
where,
V = Electric potential
E = Electric field
r = Distance from the plate

Solving for the electric field:
E = \frac{V}{r} = \frac{12}{0.02} = 600 V/m

The electric field, E, is related to the charge density, \sigma , on the plate by:
E = \frac{\sigma}{\epsilon_0}
where,
\sigma = Charge density
\epsilon_0 = Permittivity of free space (8.85 \times 10^{-12} \, C^2/Nm^2)

Solving for charge density:
\sigma = E \times \epsilon_0 = 600 \times 8.85 \times 10^{-12} = 5.31 \times 10^{-9} C/m^2

To find the total charge on the plate, we need to multiply the charge density by the area of the plate:

Given:
- Plate width, w = 16 cm = 0.16 m
- Plate length, l = 200 mm = 0.2 m

The area of the plate, A, is:
A = w \times l = 0.16 \times 0.2 = 0.032 \, m^2

Finally, the total charge, Q, on the plate is:
Q = \sigma \times A = 5.31 \times 10^{-9} \times 0.032 = 1.7 \times 10^{-10} \, C

\boxed{Q = 1.7 \times 10^{-10} \, C}