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?
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
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]
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]
Julie
What do u mean by + same