Why is it safer to be inside a car than standing under a tree during lightning ?

Question 79 of 122

Q. Why is it safer to be inside a car than standing under a tree during lightning ?

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 3mImportance★★★★★

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The metal body of the car behaves as a Faraday cage, shielding the interior from the lightning's electric field, while a tree offers no such protection and can even discharge current onto a person standing beneath it.

A car's outer body is essentially a closed (or nearly closed) hollow metallic conductor. If lightning strikes the car, the induced charge distributes itself over the outer metal surface of the body, exactly like any charged hollow conductor. Because the electric field inside a charged conductor is always zero (all the charge resides on the outer surface, as in Q24 above), the strong electric field and the surge of current associated with the strike are confined to the outer skin of the car and are safely conducted to the ground through the body and tyres, without significantly penetrating the passenger compartment. The occupants inside, being enclosed by this conducting shell, are therefore electrostatically shielded from the lightning's field — this is the same principle as a Faraday cage.

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set 55/4/11 markMCQ
$q_1$
›Reveal solutionSolution
Because $q_2$ sits at the centre of its own conducting shell, that shell electrostatically shields $q_2$ from any external field — including the field of $q_1$ and its shell. The force on $q_2$ due to $q_1$ is $\boxed{\text{Zero}}$ , option (C).

The core concept here is electrostatic shielding: the field inside the cavity of a conductor depends only on the charge enclosed within that cavity, never on anything outside the conductor — however strong, however close. This is exactly the property that makes a Faraday cage work, and NCERT states it explicitly for the cavity-in-a-conductor case.

Important

It is tempting to treat each shell-plus-central-charge system as an effective point charge (true for the field outside an isolated shell) and then simply plug into Coulomb's law. That reasoning correctly describes the field outside shell 1 at large distances. But it silently assumes $q_2$ is sitting bare in that field — it is not. $q_2$ is enclosed within its own conducting shell (shell 2), and that changes everything about the force actually experienced by $q_2$ .

Why the field cannot reach $q_2$

Shell 2 is a conductor in electrostatic equilibrium, so the field is zero everywhere within the metal of shell 2 itself.

The charge on the inner surface of shell 2 is fixed by $q_2$ alone. Because $q_2$ sits exactly at the centre of the spherical cavity, the induced charge on the inner wall must be a uniform layer of total charge $-q_2$ — this is the only distribution consistent with (a) the field being zero in the conductor bulk, and (b) $q_2$ being at the geometric centre. This uniform layer is completely determined by $q_2$ 's position; it has nothing to do with $q_1$ or shell 1.

Whatever the external field from shell 1 does, it can only rearrange the outer surface charge of shell 2 (making it non-uniform, so that shell 2's outer surface, seen from far away, is no longer a perfect point-charge equivalent). This outer-surface rearrangement is exactly what "soaks up" the external field — it cannot communicate anything back through the conducting bulk to the inner surface or the cavity. …
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Q.Why is it safer to be inside a car than standing under a tree during lightning ?

Concept understanding — Electrostatic Shielding

Why the field cannot reach $q_2$

Q.Why is it safer to be inside a car than standing under a tree during lightning ?

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