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Why electric field inside a hollow sphere on points except centre is 0

I got the explation through the gauss law but I can mannually see at points except centre the field is non zero

Qmechanic
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1 Answers1

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Consider a Gaussian surface inside the hollow conducting sphere:

Choose a spherical Gaussian surface with radius ( r ) (where ( r ) is less than the inner radius of the hollow sphere) and center at the center of the sphere. By symmetry, the electric field ( E ) should be the same at every point on this Gaussian surface, pointing radially outward (or inward, but consistently in one direction). Since the Gaussian surface encloses no charge (because all charges are on the outer surface of the hollow sphere), Gauss's Law gives us:

$$ \oint_{\text{Gaussian surface}} E \cdot dA = \frac{Q_{\text{enc}}}{\epsilon_0} $$

Here,$$ \left( Q_{\text{enc}} = 0 \right) $$ because there is no charge inside the hollow region:

$$ \oint_{\text{Gaussian surface}} E \cdot dA = 0 $$

Given that the electric field ( E ) is constant on the Gaussian surface, the above integral simplifies to:

$$ E \cdot 4\pi r^2 = 0 $$

Therefore,

$$ E = 0 $$

This shows that the electric field ( E ) must be zero everywhere inside the hollow sphere.

Addressing the Intuition of Non-Zero Fields at Points Other Than the Center: It might seem intuitive to think that at points other than the center, the electric field should be non-zero due to the contributions from different parts of the charged surface. However, the uniformity and symmetry of the charge distribution mean that these contributions perfectly cancel out at every point inside the hollow sphere, not just at the center.

If you were to consider the vector sum of all the infinitesimal electric fields from the surface charges at any point inside the sphere, these vectors would cancel each other out due to symmetry. Thus, the resultant electric field is zero at every point inside the hollow sphere, maintaining the condition predicted by Gauss's Law.

Stefy
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