Question

The conductors are separated by a distance of
1cm in air. The dielectric strength of air is about
3 X 106V m 1. What maximum safe potential
difference can be applied across conductors
(A) 2 x 104V
(B) 6 x 1047​

Answer 1

Answer:

Correct option is

D

2122 V

Given : E=3×10

6

V/m d=0.001 m

Thus maximum r.m.s voltage V

max

=

2

Ed

=

2

3×10

6

×0.001

⟹ V

max

=2122 V

Explanation:

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Answer 2

Given:

Distance = 1cm

Dielectric constant = 3 × 10⁶ m⁻¹

To find:

Maximum safe potential difference

Solution:

When the conductors are connected, then their maximum safe potential difference applied across conductors will be calculated as -

  VD(max.) = 2 × d × k

here, VD(max.) = maximum potential difference

        d = distance between the conductors

        k = dielectric constant

convert distance into meters,

d = 1cm = 0.01m

k = 3 × 10⁶  m⁻¹

by substituting the values, we get

VD(max.) = 2 × 0.01 × 3 × 10⁶

                = 0.06 × 10⁶ V

So, the maximum potential difference applied across conductors= 6 × 10⁴ V.