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Chapter 23--Examples. -2q. -3q. +5q. d. d. d. d. P. +3q. d. d. +5q. -2q. Problem. In the figure below, point P is at the center of the rectangle. With V=0 at infinity, what is the net electric potential at P due to the six charged particles?. s. d/2. d. -2q. -3q. +5q. d. d.

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Chapter 23--Examples


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problem

-2q

-3q

+5q

d

d

d

d

P

+3q

d

d

+5q

-2q

Problem
  • In the figure below, point P is at the center of the rectangle. With V=0 at infinity, what is the net electric potential at P due to the six charged particles?
find distance from corners to p

s

d/2

d

-2q

-3q

+5q

d

d

d

d

P

+3q

d

d

+5q

-2q

Find distance from corners to P
potentials voltage is a scalar

s

d/2

d

-2q

-3q

+5q

d

d

d

d

P

+3q

d

d

+5q

-2q

Potentials (Voltage) is a scalar

2

3

1

4

5

6

problem6
Problem

A charge q is distributed uniformly throughout a spherical volume of radius, R.

  • Setting V=0 at infinity shown that the potential at a distance r from the center, where r<R is given by
  • What is the potential difference between a point on the surface and the sphere’s center?
okay that is if v 0 at infinity what if v 0 at the center of the sphere
Okay, that is if V=0 at infinity what if V=0 at the center of the sphere?

Same as previous

Same as previous

problem13
Problem

The electric potential at points in a space are given by

V=2x2-3y2+5z3

What is the magnitude and direction of the electric field at the point (3,2,-1)?

directions
Directions

Direction w.r.t +x axis

Direction w.r.t +z axis

problem16
Problem

Three +0.12 C charges form an equilateral triangle, 1.7 m on a side. Using energy that is supplied at a rate of 0.83 kW, how many days would be required to move one of the charges to the midpoint of the line joining the other two charges?

draw it
Draw It

Initially, this charge is 1.7 m from the other two charges

0.12C

1.7 m

1.7 m

0.85 m

0.85 m

0.12C

0.12C

1.7 m

Finally, this charge is 0.85 m from the other two charges

power work per unit time
Power=Work per unit time
  • P=W/Dt
  • W=-DU
  • So 0.83 kW= 830 J/s
  • And Dt= DU/P=152x106/830
  • Dt=183,699 s or 51 hours or 2.12 days