Aparticle of charge 3.53×10 −8 c experiences a force of magnitude 6.03×10 −6 n when it is placed in a particular point in an electric field. what is the magnitude of the electric field at that point?

Explanation:

Electric field is the ratio of force and charge.

Force, F = 6 x 10⁻⁶ N

Charge, q = 3.53 x 10⁻⁸ C

We have           

Electric field at the location of the charge is 169.97 N/C

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm