Question

A particle of mass 5 x 10 to the power minus 6 gram is kept over charge horizontal sheet of charge density 4 x 10 to the power minus 6 coulomb per metre square what charge should be given to this particle so that is released it does not fall down how many lactone are to be removed to give the charge

Answer 1

Answer:

We know that the electric field E for a sheet is  E=σ/2ε where 2ε is due to both side of the sheets.

Where σ is the given charge density 4*10^-6 C/m^2.

So, we also know that the value of force F=qE where we can calculate the value of the charge q will be 5×10^−6×9.8×2×8.85×10^−12/4×10−6 where we already know the value of g =9.8 m/s^2 mass is given as 5×10^−6 and the value of ε is 8.85×10^−12.

So, on solving the equation we will get the value of q as 2.16*10^-10 Coulomb.

Answer 2

Answer:

answer is in the above image

Explanation:

hope it help u