Question

Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106V/m ) if the plates are separated by 2.00 mm and a potential difference of 5.0×103V is applied? (b) How close together can the plates be with this applied voltage?

Answer 1

(a) The electric field strength between two parallel conducting plates does not exceed the breakdown strength for air ($3 \times 10^{6} V / m$)

(b) The plates can be close together to 1.7 mm with this applied voltage

Explanation:

Given data:

Dielectric strength of air = $3 \times 10^{6} V / m$

Distance between the plates = 2.00 mm = $2.00 \times 10^{-3} \mathrm{m}$

Potential difference, V = $5.0 \times 10^{3} V$

We need to find

a) whether the electric field strength between two parallel conducting plates exceed the breakdown strength for air or not

b) the minimum distance at which the plates can be close together with this applied voltage.

The voltage difference (V) between two points would be equal to the product of electric field (E) and distance separation (d). The equation form is and apply all given value,

         $E=\frac{V}{d}=\frac{5.0 \times 10^{3}}{2.00 \times 10^{-3}}=2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$

From the above, concluding that The electric field strength between two parallel conducting plates ($2.5 \times 10^{6} \mathrm{V} / \mathrm{m}$) does not exceed the breakdown strength for air ($3 \times 10^{6} V / m$)

b) To find how close together can the plates be with this applied voltage:

The formula would be,

            $d_{\min }=\frac{V}{E_{\max }}$

Apply all known values, we get

      $d_{\min }=\frac{5.0 \times 10^{3}}{3 \times 10^{6}}=1.7 \times 10^{-3} \mathrm{m}=1.7 \mathrm{mm}$