Задачи по "Физике"

[pic 1]a[pic 2]

                                                -        [pic 3]       -

Задача 1.

Дано:

q=3 мкКл

а=10см=0,1м

Найти:

FA=?

φB=?

AAB=?

Решение: 1.α=45̊ (бок. стороны равны)

FA=[pic 4] (по теореме косинусов)

[pic 5]

[pic 6][pic 7]

                                FA                                                        

FAB

                        FAC                α        

FAB=[pic 8]

FCA=[pic 9]

FA=[pic 10]

FA=0.75*[pic 11]

2.принцип Суперпозиции :

φ = [pic 12]

или

φB=|[pic 13]|=φ[pic 14]BB+ φ[pic 15]BA

φB=[pic 16]

φBC=[pic 17]

φBA=[pic 18]

φB=[pic 19]

φB

[pic 20][pic 21][pic 22]

                        φC

φA[pic 23]

                α        B[pic 24][pic 25]

                C                 A

3.

+ A

S                A[pic 26][pic 27]

C -                        - B

A=[pic 28]Wп=WB –WA

[pic 29]Wп=[pic 30]

R1=r2=a=>   [pic 31]Wп=0

          =>  A=0

2)

                                                    +     I1                              I3                                                       I3[pic 32][pic 33][pic 34][pic 35][pic 36][pic 37][pic 38][pic 39][pic 40][pic 41][pic 42][pic 43][pic 44][pic 45][pic 46][pic 47][pic 48][pic 49][pic 50][pic 51][pic 52][pic 53][pic 54][pic 55][pic 56]