Question

the surface density on the copper sphere is sigma. what will be the electric field strength on the surface of the spheres

Answer 1

Answer:

• Electric field strength (E) is σ/ε₀

Given:

1. Surface charge density is σ

Explanation:

$\rule{300}{1.5}$

From the Gauss law we know,

$\large\bigstar\;\underline{\boxed{\sf \Delta\;\phi=\displaystyle\oint\sf E.da=\dfrac{Q}{\in_{0}}}}$

Here,

• ΔФ denotes flux.
• E Denotes electric field.
• dA Denotes elemental portion of area.
• Q Denotes the charge enclosed.
• ε₀ Denotes permittivity of free space.

It is given that surface charge density of sphere is σ, it can be written as,

$\longmapsto\sf \sigma=\dfrac{Q}{4\pi R^{2}}$

        OR

$\longmapsto\sf \sigma\times 4\pi R^{2}=Q\\\\\\\longmapsto{\underline{\underline{\sf Q=\sigma\;.\; 4\pi R^{2}}}}\sf \dots\dots(1)$

(Where R is the radius of the sphere.)

Now solving it through Gauss law,

$\longmapsto \displaystyle\oint\sf E.dA=\dfrac{Q}{\in_{0}}\\\\\\\\\longmapsto\sf E\displaystyle\oint\sf dA=\dfrac{Q}{\in_{0}}\\\\\\\\\longmapsto\sf E\times 4\pi R^{2}=\dfrac{Q}{\in_{0}}\ \ \ \ \ \because\Bigg[\displaystyle\oint\sf dA=4\pi R^{2}\Bigg]\\\\\\\\\longmapsto\sf E=\dfrac{Q}{4\pi R^{2}.\in_{0}}$

• Substituting the value of Q from equation (1)

$\longmapsto\sf E=\dfrac{Q}{4\pi R^{2}.\in_{0}}\\\\\\\\\longmapsto\sf E=\dfrac{\sigma\;.\; 4\pi R^{2}}{4\pi R^{2}.\in_{0}}\\\\\\\\\longmapsto\sf E=\dfrac{\sigma\;.\; \not4\not\pi \not R^{2}}{\not4\not\pi \not R^{2}.\in_{0}}\\\\\\\\\longmapsto\sf E=\dfrac{\sigma}{\in_{0}}\\\\\\\\\longmapsto\large{\underline{\boxed{\red{\sf E=\dfrac{\sigma}{\in_{0}}}}}}$

∴ Electric field strength (E) is σ/ε₀.

$\rule{300}{1.5}$

Answer 2

Answer:

0/3° is the correct answer

Explanation:

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