Question

Two equal point charges of 5 µC are on the horizontal axis. One charge is at -10 cm, the other is at +10 cm.?

These two charges don't move, they are STUCK, take the potential to be zero at infinity.
Find the potential at the following coordinates: (0,0) cm, (0,10)cm & (0,20)cm.
What will happen to a proton released from rest at the origin?

Answer 1

The potential at any point is the sum of the potentials due to the two separate charges. (that is the thing about potentials, they add)
due to charge 1
k = 1/(4 pi eo)
Q = 5 u C

U1 = k Q /((x+.1)^2+y^2)^.5
due to charge 2
U2 = k Q /[( x-.1)^2 +y^2) ]^.5
so
at (0,0)
U = U1 + U2 = k Q [1/.1^2+1/.1)^2

at (0,10)
U = U1+U2= k Q [1/(.1^2+.1^2)^.5 + same]

at (0,20)
k Q [ 1/(.1^2+.2^2)^.5 + same]

Answer 2

What do u mean by + same