Question

The electric fields in which a charged oil deop of mass 5mg and charge 10C be held stationary is

Answer 1

Answer:

E = 4.9 ×$10^{-7}$ N/C

Explanation:

To keep the oil drop stationary, the force applied by the electric field must be equal to the force of gravitation on the oil drop, that is pulling it down.

Force of gravitation on oil drop ($F_g$) = mg

"           applied by the electric field of magnitude (E), $(F_E)$ = qE

So,

($F_g$)  = $(F_E)$

or mg = qE

or E = mg/q

or E = $\frac{5*10^{-7}*9.8}{10}$

∴ E = 4.9 ×$10^{-7}$ N/C

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