Question

Find the value of the electric intensity at a point at a distance of 15 cm in air from
a.Point charge of 450 µc.

Answer 1

Answer:

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Answer 2

Electric field will be $2.7\times 10^7N/C$

Explanation:

Here it is given electric intensity i think we have to find electric field intensity

It is given distance of the point where we have to find the electric field r = 15 cm = 0.15 m

Point charge $q=450\mu C=450\times 10^{-6}C$

Electric field intensity is given by

$E=\frac{1}{4\pi \epsilon _0}\frac{q}{r}$

So electric field will be $E=\frac{1}{4\pi \epsilon _0}\frac{q}{r}=\frac{9\times 10^9\times 450\times 10^{-6}}{0.15}=2.7\times 10^7N/C$

Learn more

Electric Field and Electric Field Lines

1. The point charges of q and 4q are kept 30 cm apart. At a distance on the straigh

line joining them, intensity of electric field is zero.

(A) 20 cm from 4q

(B) 7.5 cm from a

(C) 15 cm from 4g

(D) 5 cm from a​

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