Physics, asked by srushtim08, 10 months ago


47) How many electron must be removed from the
I metal sphere of ladies 0.05 m, so that it acquires
a charge of 4 x 10-15 c ? What is the electric field at
appoint (i) on the surface of the Sphere ii) at a
point 0.04 m from the centre of the Sphere?
(ec=1.6X.1 Ö-19 C).​

Answers

Answered by Kul1306
3

Answer:

Charge of each electron e=1.6×10^−19C

Charge of each electron e=1.6×10^−19CNumber of electrons removed from the sphere n=4×10^10

Charge of each electron e=1.6×10^−19CNumber of electrons removed from the sphere n=4×10^10Thus total charged removed from the sphere q=ne

Charge of each electron e=1.6×10^−19CNumber of electrons removed from the sphere n=4×10^10Thus total charged removed from the sphere q=ne∴ q=4×10^10×1.6×10^−19=6.4×10^−9C 

Charge of each electron e=1.6×10^−19CNumber of electrons removed from the sphere n=4×10^10Thus total charged removed from the sphere q=ne∴ q=4×10^10×1.6×10^−19=6.4×10^−9C Distance of point from the center of sphere r=20cm=0.2m

Charge of each electron e=1.6×10^−19CNumber of electrons removed from the sphere n=4×10^10Thus total charged removed from the sphere q=ne∴ q=4×10^10×1.6×10^−19=6.4×10^−9C Distance of point from the center of sphere r=20cm=0.2mThus electric field at that point E=kq/r^2  

Charge of each electron e=1.6×10^−19CNumber of electrons removed from the sphere n=4×10^10Thus total charged removed from the sphere q=ne∴ q=4×10^10×1.6×10^−19=6.4×10^−9C Distance of point from the center of sphere r=20cm=0.2mThus electric field at that point E=kq/r^2  where k=1/4πϵo1=9×10^9

Charge of each electron e=1.6×10^−19CNumber of electrons removed from the sphere n=4×10^10Thus total charged removed from the sphere q=ne∴ q=4×10^10×1.6×10^−19=6.4×10^−9C Distance of point from the center of sphere r=20cm=0.2mThus electric field at that point E=kq/r^2  where k=1/4πϵo1=9×10^9⟹ E=9×109×6.4×10^−9/ (0.2)^2=1440NC^-1

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