Suppose that in Figure P22.8, |q2| > |q1| > 0 hut these par- tides may carry cha

Question

Suppose that in Figure P22.8, |q2| > |q1| > 0 hut these par- tides may carry charges of the same or opposite type. A third particle carrying charge q3 is to be placed in this system along the axis such that the vector sum of electric forces on the third particle is zero. In which region I, II, or III might this be possible if (a) q1 and q2 have the same sign and (b) q1 and q2 have opposite signs? (c) Under what circumstances can the third particle be placed in region I and have the vector sum of electric forces exerted on it be zero?..

Explanation / Answer

We use the relationship that charge of same sign repel each other and charge of different sign attract each other

Using the second Newton lay

F32 + F12 = 0

The only way this happens is that F13 and F23 have opposite directions. I don't see your figure, but we will try to answer.

We assume that q3 is positive charge (the most common case)

Part a

Region   I II                                                    III

.           + q3            +q2 + q1       

                     <------- F23

<-- F12                  - F23 - F13 <0

Region II +q2                          +q3            +q1

-----> F23

<------- F13       F23-F13 = 0    F23= F13

                                                                       .         k q2 q3/ r232 = k   q1 q3/ r132

                                                    .       q2/q1 = (r23/r13)2  

It must be met so that the net force is zero It must be met so that the net force is zero

In this area there is a point in the II region (between the two charges) where the sum of the forces are zero

       Region III                 .   +q2                         +q1               +q3

                                                                                    ----------> F13

                                                                                    ----> F23       F13+F23 >0

Part b

Region I           +q3                            +q2                      -q1

                                       <---------     F23

                                              ----> F13                   -F23 + F13 = 0

Following the same development to find the condition of neutrality

    .     k q2q3/r232 = k q1 q3/ r132       (r13/r23)2 = q1/q2

Region II            +q2            +q3                  -q1

                                    ------> F23

                                       --> F13   F23+F13 > 0

Region III       +q2 -q1               +q3

                                                                             <----------    F13

                                                                             ----> F23                 F23-F13 = 0

Cero force point       (r13/r23)2 = q1/q2

Neutral point
the most common case
To meet the relation of neutrality depends on the value of the charges

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