grav and electric fields exam style qs

Question 1

Answer 1

equate ke = pe

1/2mv^2=mgh

GM/r = 1/2v^2

potential = 1/2v^2

re arrange for v and sub in potential at surface = 7.4x10^5

this will get min speed of 1200 ms^1

1400>1200 so it can escape

Question 2

Define the electric field strength at a point in an electric field.

Answer 2

force per unit charge on a (small) positive charge (at that point)

Question 3

Answer 3

(At B) the (magnitude) of the electric field strength due to Q = the magnitude of the electric field strength due to the 46 μC charge

𝑄 = 6.9 x 10^–5 (C) (68.7 μC rounding must be correct)