Each plate of a parallel-plate air-filled capacitor has an area of 2×10−3 m2, and the separation of the plates is 5×10−2 mm. an electric field of 8.5 ×106 v/m is present between the plates. what is the surface charge density on the plates? (ε 0 = 8.85 × 10-12 c2/n · m2)

The surface charge density on the plate,

Explanation:

It is given that,

Area of parallel plate capacitor,

Separation between the plates,

Electric field between the plates,

We need to find the surface charge density on the plates. The formula for electric field is given by :

Where

= surface charge density

Hence, this is the required solution.

i believe the answer would be:

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