CK-12-Physics - Intermediate

16.2. Electric Potential http://www.ck12.org

V=Ex=

(

1. 30 × 105

V

m

)

( 8. 00 × 10 −^3 m) = 1. 04 × 103 V

e. How much work has the field done on the particle as it moved from one plate to the other?

W=q∆V= ( 3. 00 × 10 −^3 C)( 1. 04 × 103 V) = 3. 12 J

Illustrative Example 16.2.3

An electron is accelerated from rest through a potential difference of 30,000 V. The mass of the electron is 9. 11 ×
10 −^31 kgand the charge of the electron is 1. 60 × 10 −^19 C. Find its velocity.

Answer: Recall that the Work-Energy Principle states thatW=∆KE.

W=∆KE

W=q∆V

∆KE=q∆V

1

2

mv^2 f−

1

2

mv^2 i=q∆V

vi= 0 →v^2 f=

2 q∆V

m

→vf=

√

2 q∆V

m

→

vf=

√

2 ( 1. 60 × 10 −^19 C)( 3. 00 × 104 V)

9. 11 × 10 −^31 kg

= 1. 026 × 108 → 1. 03 × 108

m

s

Illustrative Example 16.2.4

What magnitude of an electric field is required to balance the gravitational force acting on an electron inFigure
16.7?

FIGURE 16.7

Illustrative Example 16.2.4-An electron suspended in an electric field.

Answer:

Draw a Free-Body-Diagram (FBD) of the situation. The electrostatic force that acts on the electron points upward
and the gravitational force that acts upon on the electron points downward. The electron is suspended motionless (or
moves with a constant velocity) when the net force on the electron is zero.