Question

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of (1.50 X 10⁴ V)?

a) (1.50 X 10² V/m)
b) (1.50 X 10³ V/m)
c) (1.50 X 10⁴ V/m)
d) (1.50 X 10⁵ V/m)

Answer 1

The strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference of 1.50 × 10⁴ V is 1.50 × 10⁶ V/m.

Step-by-step explanation:

The strength of the electric field between two parallel conducting plates can be calculated using the formula:

Electric field strength (E) = Voltage (V) / Separation distance (d)

Given that the potential difference between the plates is 1.50 × 104 V and the separation distance is 1.00 cm, we can substitute these values into the formula to find the electric field strength:

E = (1.50 × 104 V) / (1.00 cm)

Convert the separation distance to meters:

E = (1.50 × 104 V) / (0.01 m) = 1.50 × 106 V/m

Therefore, the strength of the electric field between the two parallel conducting plates is 1.50 × 106 V/m.