Introductory Physics Volume Two

50 Electric Potential 2.7
The change in potential energy of the electron will be:
Use conservation of energy:
∆U = q∆V = (−e)∆V.
∆U + ∆K = −e∆V + ( 1 2 m ev 2 − 0) = 0
√ √
2e∆V 2(1.6 × 10
−→ v = =
−19 C(+9V)
m e 9.11 × 10 −31 kg
= 1.78 × 10 6 m s
Example
Near the surface of the Earth, there is always a background electric
field that averages 100N/C and points down. Assuming the Earth to be
a perfect conductor, how much electric charge is stored on the Earth’s
surface?
Since the electric field points down, we know the Earth’s charge is negative.
So, lets find the magnitude. Use the previous result to determine
the surface charge density:
E = σ ɛ 0
−→ σ = Eɛ 0 .
Multiply this by the Earth’s surface area to get the total charge:
Q E = σ(4πR 2 E) = Eɛ 0 (4πR 2 E) = 453, 000C
Example
Two equal and opposite charges are located a distance a apart. Find
the electric field and the electric potential at a) the point directly between
the two charges and b) the point a distance a/2 directly above
the point inbetween the two charges.
+q -q
The electric fields due to each charge point in the same direction at the
center point c:
a
E +
c
+q -q
E -