Question

Two conductors are separated by distance
of 1 cm in air. The dielectric strength of air is
about 3 106 Vm-1. What maximum safe
potential difference can be applied across
conductors?
​

Answer 1

Answer:

The maximum safe  potential difference can be applied across conductors is 31.06 Volts.

Explanation:

Given, distance of separation(d)= 1cm= $10^{-2}$m.

Dielectric strength of air (K)= 3106 $Vm^{-1}$

Voltage to be applied= K×d= 3106×$10^{-2}$=31.06 Volts.

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Answer 2

Given: the dielectric strength of air is about 3 x 10^6 V/m.

To find: The maximum safe potential difference that can be applied across two conductors separated by a distance of 1 cm in air.

Solution:

Using the formula for electric field strength E = V/d, where E is the electric field strength, V is the potential difference, and d is the distance between the conductors.

Substituting the values, we get E = (V/0.01) = 3 x 10^6 V/m.

Rearranging the formula, we have V = E x d = (3 x 10^6 V/m) x 0.01 m = 30,000 V.

Therefore, the maximum safe potential difference that can be applied across two conductors separated by a distance of 1 cm in air is 30,000 V.

To learn more about conductors from the given link.

https://brainly.in/question/15510573

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