An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by a distance of 1.90 mm. If a 18.2-V potential difference is applied to these plates, calculate the following. (a) the electric field between the plates -9.6 x Apply the relation between the field strength and the potential difference between two points in a uniform electric field. kv/m ---Select--- magnitude direction (b) the capacitance DF (c) the charge on each plate pC
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An electrostatic force is a force caused by stationary electric charges /fields. The electrostatic force is caused by the transfer of electrons in conducting materials. Coulomb’s law determines the amount of force between two stationary, charged particles. The electric force is the force which acts between two stationary charges. It is also called Coulomb force.
![An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm², separated by a distance of 1.90 mm. If a 18.2-V potential difference is applied to these plates, calculate the following:
(a) The electric field between the plates
- **Magnitude:** [Input Box] “-9.6” [Red Cross]
- **Guidance Message:** Apply the relation between the field strength and the potential difference between two points in a uniform electric field. (kV/m)
- **Direction:** [Dropdown Menu: ---Select---]
(b) The capacitance
- [Input Box] pF
(c) The charge on each plate
- [Input Box] pC
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Data provided:
An air filled parallel plate capacitor
Area (A) = 7.60 cm2 = 7.60×10-4 m2
Distance (d) = 1.90 mm = 1.90×10-3 m
Potential difference (V) = 18.2 V
Asked:
The electric field, capacitance and charge on each plate
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