Question

A 2.5-cm-diameter parallel-plate capacitor has a 2.2 mm spacing. The electric field strength inside the capacitor is 6.0×104 V/m . You may want to review (Pages 699 - 702) . Part APart complete What is the potential difference across the capacitor? Express your answer to two significant figures and include the appropriate units.

Answer 1

Explanation:

The given data is as follows.

     Separation between two plates, d = 2.2 mm = $2.2 \times 10^{-3} m$

     Electric field strength inside the capacitor, E = $6.0 \times 10^{4} V/m$

Therefore, we will calculate the potential difference across the capacitor as follows.

              V = Ed

                  = $6.0 \times 10^{4} V/m \times 2.2 \times 10^{-3} m$

                  = 1.32 V

Thus, we can conclude that the potential difference across the given capacitor is 1.32 V.

Answer 2

The potential difference across the capacitor is 132 Volts.

Capacitor

A capacitor is an electrical device that stores energy in its electrical field. The potential difference (V) between the flat plate of a capacitor is given by:

V = Ed

where E is electric field strength = 6.0 × 10⁴ V/m, d is the distance = 2.2 mm = 0.0022 m. Hence:

V = 6.0 × 10⁴ * 0.0022 = 132 Volts

The potential difference across the capacitor is 132 Volts.

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