Suppose you have two parallel couducting plates that are separated by 2.7 mm. Wh

Question

Suppose you have two parallel couducting plates that are separated by 2.7 mm. What will the electric field strength between the plates be (in N/C) if they have a difference of 4.8 Times 10^3 V? The electric breakdown strength for a particular medium, also called the dielectric strength, is the point at which electron bound to the molecules of the medium begin to be stripped off due to the large electric field. How close together must the plates be with this applied voltage in order to achieve breakdown strength for air (3.0 Times 10^6 V/m) in mm?

Explanation / Answer

Part (a): Distance between the parallel plates, d = 2.7 mm = 2.7 * 10^-3 m

Now, E = V/d , Where E is the electric field strength between the parallel plates.

=> E = 4.8 * 10^3/2.7 * 10^-3 = 1.78 * 10^6 V/m = 1.78 * 10^3 kV/m

Part (b): Breakdown strength for air = 3.0 * 10^6 V/m

So, the distance in this case is 4.8 * 10^3/3.0 * 10^6 = 1.6 * 10^-3 = 1.6 mm

Therefore, the distance between the parallel plates must be 1.6 mm to achieve the electric breakdown.

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