CK-12-Physics - Intermediate

http://www.ck12.org Chapter 16. Electric Potential

Illustrative Example 16.2.2

a. A particle of massmof 2. 00 × 10 −^5 kgis has a chargeqof+ 3. 00 × 10 −^3 C. If the particle is released from the
positive plate of a parallel-plate conductor with an electric fieldEof 1. 30 × 105 NC, determine the acceleration of the
particle, seeFigure16.6.

FIGURE 16.6

Illustrative Example 16.3.2

Answer:

If we ignore gravity, the only force acting on the particle is the electric forceF=qE. Using Newton’s Second Law,
the net force on the particle is equal to∑F=ma→qE=ma.

The acceleration isa=Eqm=(^1.^30 ×^10

5 Vm)( 3. 00 × 10 − (^3) C)
2. 00 × 10 −^5 kg =^1.^95 ×^10
7 V∗C
kg∗m.
b. Show that the unitskgV∗∗Cmare equivalent to the unitssm 2.
Answer:

V∗C

kg∗m

=

J

kg∗m

=

N∗m

kg∗m

=

N

kg

=

kg∗ms 2

kg

=

m

s^2

c. The plates have separation of 8.00 mm. Determine the velocity of the particle when it reaches the negative plate

Answer:

This is a kinematics problem, where the displacement and acceleration are known and the velocity is to be found.
Recall the equationv^2 f=v^2 i+ 2 a∆x.

→v^2 f= 0 + 2

(

1. 95 × 107

m

s^2

)

( 8. 00 × 10 −^3 m) = 312 , 000

m^2

s^2

v= 558. 6 → 5. 59 × 102

m

s

.

d. What is the potential difference between the plates?

Answer: