Question

Deduce the electric field intensity
due to a point charge. Hence Calculate
the force experienced due to test
charge?​

Answer 1

Answer:

$\huge\mathfrak\red{Answer :-}$

Electric field intensity is the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge placed at that point, or

E = F/q

From Gauss law, considering a spherical surface with charge at its center as Gaussian surface,

E(A) = q/e

$⟹ \: E(4\pi \: {r) }^{2} = \frac{q}{e}$

$⟹ \: E = \frac{q}{4\pi \: e {r}^{2} }$

which is the intensity of electric field at a distance r from the charge.

Explanation:

Hope this may help you....

Answer 2

Answer:-

Electric field intensity is the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge placed at that point, or

E = F/q

From Gauss law, considering a spherical surface with charge at its center as Gaussian surface,

E(A) = q/e

⟹ \: E(4\pi \: {r) }^{2} = \frac{q}{e}⟹E(4πr)

2

=

e

q

⟹ \: E = \frac{q}{4\pi \: e {r}^{2} }⟹E=

4πer

2

q

which is the intensity of electric field at a distance r from the charge.

Explanation:

hope it's helpful to you..

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