Lightning Can Hurt You

Electricity and Magnetism Level 1

Lightning occurs when there is a breakdown in the electrical insulating properties of air. That is, the potential difference between the ground and a cloud is so great that it overcomes the fact that air usually a pretty good insulator. Air begins to conduct at a potential difference of approximately $3 \times 10^6\text{ V/m}$ .

A typical height for a cloud to ground lightning strike is 1 km.

What is the total voltage difference between the cloud and the ground right before the lightning strike in Volts ?

Image credit: Wikipedia

The answer is 3E+9.

3 solutions

Discussions for this problem are now closed

Prasun Biswas
Mar 15, 2014

Given that-----

Potential difference per unit length on travel of charges $(E)=3\times 10^6 V/m$ and the length through which the charges of the cloud are travelling from cloud to ground (say $L) = 1\text{ km}=10^3 \text{m}$ .

Now, we know that----

$\text{ (Total voltage difference) }=\text{ (Potential difference) } \times \text{ (Distance covered by the electric charges) }$

$\implies V=EL$

$\implies V=(3\times 10^6 \text{V/m})\times 10^3\text{ m }$

$\implies V=3\times 10^9 \text{V} = \boxed{(3E+9)\text{ V}}$

Can you explain the statement : (Total voltage difference)= (Potential difference) x (Distance covered by charges)

Vijay Raghavan - 7 years, 2 months ago

Actually the given value is 3 x 10^6 V/m. Note that the unit is "V/m" which is not voltage, but actually is Electric field intensity (E). Now we know that for Potential difference (V) between the two plates with Electric Field Intensity (E) and separation (d) is given by the relation: ( V = Ed ) So the solution provided by Mr. Prasun Biswas is correct.

Sajid Ali - 7 years, 2 months ago

What I meant to say was : The above solution doesn't point out the distinction between "total voltage difference" and "potential difference".

Vijay Raghavan - 7 years, 2 months ago

hoe field equal to potential difference

Rao Rajput - 7 years, 2 months ago

potential difference is stated in terms of meters and the distance was given .

Sunetra Ganguly - 7 years, 2 months ago

it is said that the potential difference for unit distance is 3x10^6. therefore for distance 1 km or 10^3 metres will be = 3x10^6 x 10^3 = 3 x 10^9 volts

shuvam jaiswal - 7 years ago

How (3E+9)V comes

Shivani Soni - 7 years, 1 month ago

$3\times 10^9$ and $3E+9$ are different notations of the same value. So, we write the final answer as $(3E+9)V$ .

Remember that $aE+n=a\times 10^n$ where $a,b$ are $2$ real or complex nos. In this case, the values of $a,b$ are real which are $a=3,b=9$

Prasun Biswas - 7 years, 1 month ago

Vijay Raghavan
Mar 16, 2014

@David,

the question should have been

$\dots$ Air begins to conduct under the influence of an "electric field" of approximately $3\times 10^6 V/m \dots$

Now,

$V=EL$

Substituting, we get

$V= \boxed{ 3 \times 10^9 \ Volts}$

Sajid Ali
Mar 27, 2014

Actually the given value is 3 x 10^6 V/m. Note that the unit is "V/m" which is not voltage, but actually is Electric field intensity (E). Here the ground and the cloud acts as two almost parallel plates separted by some distance. Now we know that for Potential difference (V) between the two plates with Electric Field Intensity (E) and separation (d) is given by the relation: ( V = Ed ) so 3E+6 x 1000 = 3E+9 Volts

Lightning Can Hurt You

Image credit: Wikipedia

The answer is 3E+9.

3 solutions

Discussions for this problem are now closed

0 pending reports