The distance between the points a and b on two equipotential lines with v1=5.2 v and v2=1.8 v is 2.8 cm. what is the average electric field (in volt/m) at the midpoint c expressed to one decimal place?

121.142 V/m

Explanation:

A and B are on two  equipotential line.

Equipotential line are imaginary lines in space along which the potential does not change.

to find electric field at the midpoint C

we use the formula

=

=

= 121.142 V/m

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

answercollision;