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what is the strength (in v/m) of the electric field between two parallel conducting plates separated by 3.30 cm and having a potential difference (voltage) between them of 1.25 ✕ 104 v?

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Final answer:

The strength of the electric field between two parallel conducting plates, with a separation of 3.30 cm and a potential difference of 1.25 x 10^4 V, is approximately 3.79 x 10^5 V/m.

Step-by-step explanation:

To determine the strength of the electric field between two parallel conducting plates separated by a certain distance with a given potential difference, you can use the formula:

E = V/d

Where E is the electric field strength in volts per meter (V/m), V is the potential difference in volts (V), and d is the separation distance in meters (m).

In this scenario, we have a potential difference (V) of 1.25 × 104 V and a separation distance (d) of 3.30 cm, which is 0.033 m when converted to meters. Plugging these values into the formula yields:

E = (1.25 × 104 V) / (0.033 m)

After calculating, the strength of the electric field comes out to be approximately 3.79 × 105 V/m.

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