how many elecrons should be added or removed from a neutral body of mass 10mg so that it may remain hanging in air in an electric field of strength 100N/C directed upwards.

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Achleshwar Sankhla · Answer

In equilibrium F=qE =mgq=mg/E=(10×10^-3×10)/100q=0.001 CSince q=ne n=q/e =.001/1.6×10^-19 = 6.25 × 10^15

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how many elecrons should be added or removed from a neutral body of mass 10mg so that it may remain hanging in air in an electric field of strength 100N/C directed upwards.

how many elecrons should be added or removed from a neutral body of mass 10mg so that it may remain hanging in air in an electric field of strength 100N/C directed upwards.

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how many elecrons should be added or removed from a neutral body of mass 10mg so that it may remain hanging in air in an electric field of strength 100N/C directed upwards.

how many elecrons should be added or removed from a neutral body of mass 10mg so that it may remain hanging in air in an electric field of strength 100N/C directed upwards.

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