Question

The electric field strength between two parallel conducting plates separated by 4.00 cm is (7.50 X 10^4 V/m).

a) What is the potential difference between the plates?
a) (3.00 X 10^3 V)
b) (3.00 X 10^2 V)
c) (3.00 X 10^4 V)
d) (3.00 X 10^5 V)

The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?
a) (2.25 X 10^3 V)
b) (2.25 X 10^2 V)
c) (2.25 X 10^4 V)
d) (2.25 X 10^5 V)

Answer 1

The potential difference between the plates with an electric field of 7.50×104 V/m separated by 4.00 cm is 3.00×103 V. The potential at 1.00 cm from the plate with zero voltage is calculated to be 7.50×102 V, which is not included in the provided answer choices.

Step-by-step explanation:

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50×104 V/m. To find the potential difference (voltage) between the plates, you multiply the electric field strength (E) by the distance (d) between the plates:

V = E × d

Substituting the given values:

V = (7.50×104 V/m) × (4.00×10-2 m)

V = 3.00×103 V

Therefore, the potential difference between the plates is 3.00×103 V, which corresponds to answer choice (a).

For the second question, we find the potential at a point 1.00 cm from the plate with the lowest potential:

V = E × d

V = (7.50×104 V/m) × (1.00×10-2 m)

V = 7.50×102 V

The potential at 1.00 cm from the zero voltage plate is 7.50×102 V, however, this is not among the given answer choices. The potential at 1.00 cm from the zero voltage plate (and 3.00 cm from the other) should be computed like this:

V = 7.50×104 V/m × 1.00×10-2 m = 7.50×102 V

Therefore, the correct answer to the second question is 7.50×102 V, but it seems there might be a mistake in the provided answer choices.