Question: F = - ( @DIV { k@ABS { q@Sub { 0 } } @ABS { q@Sub { 3 } } ;d@Sub { 2 } @Sup

= - (

@DIV

{

k@ABS

{

q@Sub

{0}}

@ABS

{

q@Sub

{3}}

;d@Sub

{2}

@Sup

{2}})

@HAT

{

} - (

@DIV

{

k@ABS

{

q@Sub

{0}}

@ABS

{

q@Sub

{3}}

;d@Sub

{2}

@Sup

{2}})

@HAT

{

} (

Figure

3)

Now add a fourth charged particle, particle

3,

with positive charge q

3,

fixed in the yz

-

plane at

(0,

2,

2) .

What is the net force F

on particle

0

due solely to this charge?

(

kabsq

0 |

2 |

22

kabsq

0 |

1 |

12)

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