Question

obtained electric field intensity due to a point charge q = 15cm , r = 5 cm​

Answer 1

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Electric field will be 2.7 × 10^7N/C2.7×10

7

N/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}Cq=450μC=450×10

−6

C

Electric field intensity is given by

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

4πϵ

0

1

r

q

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/CE=

4πϵ

0

1

r

q

=

0.15

9×10

9

×450×10

−6

=2.7×10

7

N/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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