Advertisements
Advertisements
प्रश्न
A charged oil drop weighing 1.6 x 10-15 N is found to remain suspended in a uniform electric field of intensity 2 x 103 Nc-1. Find the charge on the drop.
Advertisements
उत्तर
Weight of the charged oil drop = `1.6 xx 10^-15`N.
Electric field = `2 xx 10^3 "NC"^-1`
∴ F = mg = qE
∴ q = `("mg")/"E" = (1.6 xx 10^-15)/(2 xx 10^3)`C
q = `0.8 xx 10^-18`C
APPEARS IN
संबंधित प्रश्न
A charge is distributed uniformly over a ring of radius 'a'. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence, show that for points at large distance from the ring, it behaves like a point charge.
The field lines of a negative point charge are as shown in the figure. Does the kinetic energy of a small negative charge increase or decrease in going from B to A?
Why must electrostatic field at the surface of a charged conductor be normal to the surface at every point? Give reason?
An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is ______.
A charge Q is placed at the centre of the line joining two point charges +q and +q as shown in the figure. The ratio of charges Q and q is ______.
Electric field at a point varies as r° for ______.
Electric lines of force about negative point charge are ______
An isolated point charge particle produces an electric field `vecE` at a point 3 m away from it. The distance of the point at which the field is `vecE/4` will be ______.
