Question

calculate the electric field due to a charge of -8×10^-8 at a distance of 5cm from it​

Answer 1

electric field due to point charge at a distance of 5 cm will be $2.88\times 10^5N/C$

Explanation:

We have given that point charge of magnitude $Q=-8\times 10^{-8}C$

We have to find the electric field at a distance of 5 cm

So $r=5cm$

As we know that 1 m = 100 cm

So $5cm=0.05m$

We know that electric filled due to point charge is given by

$E=\frac{1}{4\pi \epsilon _0}\frac{q}{r^2}$, here q is point charge, r is distance from the point charge and $\epsilon _0$ is permittivity of free space.

So electric field will be $E=\frac{Kq}{r^2}=\frac{9\times 10^9\times -8\times 10^{-8}}{(5\times 10^{-2})^2}=2.88\times 10^5N/C$

So electric field due to point charge at a distance of 5 cm will be $2.88\times 10^5N/C$

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Answer 2

Explanation:

We have given that point charge of magnitude Q=-8\times 10^{-8}CQ=−8×10

−8

C

We have to find the electric field at a distance of 5 cm

So r=5cmr=5cm

As we know that 1 m = 100 cm

So 5cm=0.05m5cm=0.05m

We know that electric filled due to point charge is given by

E=\frac{1}{4\pi \epsilon _0}\frac{q}{r^2}E=

4πϵ

0

1

r

2

q

, here q is point charge, r is distance from the point charge and \epsilon _0ϵ

0

is permittivity of free space.

So electric field will be E=\frac{Kq}{r^2}=\frac{9\times 10^9\times -8\times 10^{-8}}{(5\times 10^{-2})^2}=2.88\times 10^5N/CE=

r

2

Kq

=

(5×10

−2

)

2

9×10

9

×−8×10

−8

=2.88×10

5

N/C

So electric field due to point charge at a distance of 5 cm will be 2.88\times 10^5N/C2.88×10

5

N/C