Asked by John

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?

Answers

Answered by Damon
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]
Answered by Julie
What do u mean by + same
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions