A positive point charge q is at the center of a spherical shell of radius R carrying charge 2q distributed uniformly over its surface. Find the magnitude of the electric field 95.0 cm from the axis of the wire

a) Vb = 4.65m/s at 25.46° due south of east

b) t = 119s

c) d = 238m

d) 28.43° due north of east.

e) Vb = 3.69m/s due east

f) t = 135.5s

Explanation:

Velocity relative to the earth is:

Vb = 4.65m/s < -25.46°

Since the distance to travel is 500m:

The distance on the y-axis is given by:

If the final position is directly east from the starting position:

From the y-components:

Solving for β:

β=28.43°

With this angle, the velocity would be:

Vb = 4.2*cos(28.43°) = 3.69m/s

And the time it would take it to cross:

t = 500/3.69 = 135.5s