Questões monitoria 01 03

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FILIPE CARVALHO ANDRADE

07/04/2021

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Física III

30.183 Materiais compartilhados

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Prévia do material em texto

Q Neutra Q - q q
Inicialmente Após o contato
𝐹𝑒𝑙 =
1
4𝜋𝜀
|𝑞1||𝑞2|
𝑑2
𝐹𝑒𝑙 =
1
4𝜋𝜀
𝑄 − 𝑞 𝑞
𝑑2
𝐹𝑒𝑙 =
1
4𝜋𝜀
𝑄𝑞 − 𝑞2
𝑑2
𝑄𝑞 − 𝑞2
𝐹𝑒𝑙(𝑞) =
1
4𝜋𝜀
𝑄𝑞 − 𝑞2
𝑑2
𝐹𝑒𝑙
′ 𝑞 =
1
4𝜋𝜀 𝑑2
(𝑄 − 2𝑞)
0 =
1
4𝜋𝜀 𝑑2
(𝑄 − 2𝑞)
1
4𝜋𝜀 𝑑2
𝑄 − 2𝑞 = 0
𝑄 − 2𝑞 = 0(4𝜋𝜀 𝑑2)
𝑄 − 2𝑞 = 0
𝑄 = 2𝑞
𝑞
𝑄
=
1
2
1 2
3 4
x
y
𝑎
𝑎
𝑎
𝑎
q1 -q2
q3 -q4
x
𝑎
𝑎
𝑎
𝑎
𝑎√2
q1 -q2
q3
-q4
𝐹31
𝐹34
𝐹32
𝐹12
𝐹14
𝑎
𝑎
𝑎√2
𝐹𝑒𝑙 =
1
4𝜋𝜀
|𝑞1||𝑞2|
𝑑2
𝐹(3,1) =
1
4𝜋𝜀
200 10−9 |100 10−9 |
(5 (10−2))2
𝐹(3,1) =
1
4𝜋𝜀
2 ∗ 104 10−18
25 (10−4)
𝐹(3,1) =
1
4𝜋𝜀
2 ∗ 104 10−18
25 (10−4)
𝐹(3,1) =
1
4𝜋𝜀
2 ∗ 108 10−18
25
𝐹(3,1) =
1
4𝜋𝜀
2 ∗ 108 10−18
25
𝐹(3,1) =
1
2𝜋𝜀
1 10−10
25
𝐹(3,1) =
1
2(3,14)(8,85 ∗ 10−12)
10−10
25
𝐹(3,1) =
10−10
157(8,85 ∗ 10−12)
𝐹(3,1) =
102
1389.45
𝐹(3,1) = −0,072 𝑁
𝐹𝑒𝑙 =
1
4𝜋𝜀
|𝑞1||𝑞2|
𝑑2
𝐹(3,4) =
1
4𝜋𝜀
200 10−9 | − 200 10−9 |
(5 (10−2))2
𝐹(3,4) =
1
4𝜋𝜀
4 ∗ 104 10−18
25 (10−4)
𝐹(3,4) =
1
4𝜋𝜀
4 ∗ 104 10−18
25 (10−4)
𝐹(3,4) =
1
4𝜋𝜀
4 ∗ 108 10−18
25
𝐹(3,4) =
1
4𝜋𝜀
4 ∗ 108 10−18
25
𝐹(3,4) =
1
𝜋𝜀
1 10−10
25
𝐹(3,4) =
1
(3,14)(8,85 ∗ 10−12)
10−10
25
𝐹(3,4) =
10−10
78,5(8,85 ∗ 10−12)
𝐹(3,4) =
102
694.73
𝐹(3,4) = 0,145 𝑁
𝐹𝑒𝑙 =
1
4𝜋𝜀
|𝑞1||𝑞2|
𝑑2
𝐹(3,2) =
1
4𝜋𝜀
200 10−9 | − 100 10−9 |
(5 √2(10−2))2
𝐹(3,2) =
1
4𝜋𝜀
2 ∗ 104 10−18
25 ∗ 2 (10−4)
𝐹(3,1) =
1
4𝜋𝜀
2 ∗ 104 10−18
50 (10−4)
𝐹(3,1) =
1
4𝜋𝜀
2 ∗ 108 10−18
50
𝐹(3,2) =
1
4𝜋𝜀
2 ∗ 108 10−18
25
𝐹(3,2) =
1
2𝜋𝜀
1 10−10
50
𝐹(3,2) =
1
2(3,14)(8,85 ∗ 10−12)
10−10
50
𝐹(3,2) =
10−10
314(8,85 ∗ 10−12)
𝐹(3,2) =
102
2778.9
𝐹(3,2) = 0,036 𝑁
𝜃
tg 𝜃 =
𝐶.𝑂
𝐶.𝐴
tg 𝜃 =
𝑎
𝑎
tg 𝜃 = 1
arc tg 1= 𝜃
𝜃 = 45°
𝐹 3,2 ,𝑥 = 𝐹 3,2 ∗ cos 𝜃
𝐹 3,2 ,𝑥 = 0,036 ∗ cos 45°
𝐹 3,2 ,𝑥 = 0,036 ∗ cos 45°
𝐹 3,2 ,𝑥 = 0,025𝑁
𝐹 3,2 ,𝑦 = 𝐹 3,2 ∗ sen 𝜃
𝐹 3,2 ,𝑦 = 0,036 ∗ sen 𝜃
𝐹 3,2 ,𝑦 = 0,025𝑁
𝐹𝑥 = 𝐹 3,2 ,𝑥 + 𝐹 3,4
a)
𝐹𝑥 = 0,025 + 0,145
𝐹𝑥 = 0,17𝑁
𝐹𝑦 = 𝐹 3,2 ,𝑦 + 𝐹 3,1
b)
𝐹𝑦 = −0,047𝑁
𝐹𝑦 = 0,025 − 0,072
20𝑐𝑚 70𝑐𝑚
𝑞1 −𝑞2
20𝑐𝑚 70𝑐𝑚
𝑞1 −4𝑞1
𝐸1 =
1
4𝜋𝜀
|𝑞1|
𝑑2
𝐸1 =
1
4𝜋𝜀
|2,1 ∗ 10−8|
(𝑥 − 20 ∗ 10−2)2
𝑥
𝑥
𝐸2 =
1
4𝜋𝜀
|𝑞2|
𝑑2
𝐸2 =
1
4𝜋𝜀
| − 4 ∗ 2,1 ∗ 10−8|
(𝑥 − 70 ∗ 10−2)2
20𝑐𝑚 70𝑐𝑚
𝑞1 −4𝑞1
𝑥
𝐸 =
1
4𝜋𝜀
|𝑞2|
(𝑥 − 70 ∗ 10−2)2
−
|𝑞1|
(𝑥 − 20 ∗ 10−2)2
𝐸 = 0 𝐶𝑎𝑚𝑝𝑜 𝑁𝑢𝑙𝑜
1
4𝜋𝜀
|𝑞2|
(𝑥 − 70 ∗ 10−2)2
−
|𝑞1|
(𝑥 − 20 ∗ 10−2)2
= 0
|𝑞2|
(𝑥 − 70 ∗ 10−2)2
−
|𝑞1|
(𝑥 − 20 ∗ 10−2)2
= 0
|𝑞2|
(𝑥 − 70 ∗ 10−2)2
=
|𝑞1|
(𝑥 − 20 ∗ 10−2)2
|𝑞2|
|𝑞1|
=
(𝑥 − 70 ∗ 10−2)2
(𝑥 − 20 ∗ 10−2)2
20𝑐𝑚 70𝑐𝑚
𝑞1 −4𝑞1
𝑥
|𝑞2|
|𝑞1|
=
(𝑥 − 70 ∗ 10−2)2
(𝑥 − 20 ∗ 10−2)2
√|𝑞2|
√|𝑞1|
=
√(𝑥 − 70 ∗ 10−2)2
√(𝑥 − 20 ∗ 10−2)2
4 ∗ 2,1 ∗ 10−8
√2,1 ∗ 10−8
= √
(𝑥 − 70 ∗ 10−2)2
(𝑥 − 20 ∗ 10−2)2
±
(4 ∗ 2,1 ∗ 10−8)
√2,1 ∗ 10−8
=
(𝑥 − 70 ∗ 10−2)
(𝑥 − 20 ∗ 10−2)
±
2 2,1 ∗ 10−8
√2,1 ∗ 10−8
=
(𝑥 − 70 ∗ 10−2)
(𝑥 − 20 ∗ 10−2)
±2 =
(𝑥 − 70 ∗ 10−2)
(𝑥 − 20 ∗ 10−2)
20𝑐𝑚 70𝑐𝑚
𝑞1 −4𝑞1
𝑥
(𝑥 − 70 ∗ 10−2)
(𝑥 − 20 ∗ 10−2)
= ±2
(𝑥 − 70 ∗ 10−2)
(𝑥 − 20 ∗ 10−2)
= +2
𝑥 − 70 ∗ 10−2 = 2(𝑥 − 20 ∗ 10−2)
𝑥 − 70 ∗ 10−2 = 2𝑥 − 40 ∗ 10−2
𝑥 = −30 ∗ 10−2𝑚 𝑜𝑢 − 30𝑐𝑚
(𝑥 − 7010−2)
(𝑥 − 2010−2)
= ±2
(𝑥 − 7010−2)
(𝑥 − 2010−2)
= −2
𝑥 = 37 ∗ 10−2𝑚 𝑜𝑢 37𝑐𝑚
x
𝑎
𝑎
𝑎
𝑎q1
q2
q3q4 𝑎√2
x𝑎
𝑎
q1
-q2
q3-q4 𝑎√2
𝐸2
𝐸1
𝐸3
𝐸4
tg 𝜃 =
𝐶.𝑂
𝐶.𝐴
tg 𝜃 =
𝑎
𝑎
tg 𝜃 = 1
arc tg 1= 𝜃
𝜃 = 45°
𝑎
𝑎
q1 -q2
q3-q4
𝑎√2
2
𝐸2
𝐸1
𝐸3
𝐸4
𝐸 =
1
4𝜋𝜀
|𝑞1|
𝑑2
 𝑟
𝐸 = 𝐸1 + 𝐸2 − 𝐸3 − 𝐸4
𝐸𝑥 = 𝐸1 + 𝐸2 − 𝐸3 − 𝐸4 𝑐𝑜𝑠45°
𝐸𝑥 =
1
4𝜋𝜀𝑑2
|𝑞1| + |𝑞2| − |𝑞3| − |𝑞4| 𝑐𝑜𝑠45°
𝐸𝑥 =
1
4𝜋𝜀
|𝑞1|
𝑑2
+
1
4𝜋𝜀
|𝑞2|
𝑑2
−
1
4𝜋𝜀
|𝑞3|
𝑑2
−
1
4𝜋𝜀
|𝑞4|
𝑑2
𝑐𝑜𝑠45°
𝐸𝑥 =
1
4𝜋𝜀(
𝑎 2
2
)2
|𝑞1| + |𝑞2| − |𝑞3| − |𝑞4| 𝑐𝑜𝑠45°
𝑎
𝑎
q1 -q2
q3-q4
𝑎√2
2
𝐸2
𝐸1
𝐸3
𝐸4
𝐸𝑥 =
1
4𝜋𝜀(
𝑎 2
2
)2
|𝑞1| + |𝑞2| − |𝑞3| − |𝑞4| 𝑐𝑜𝑠45°
𝐸𝑥 =
1
4𝜋𝜀(
2𝑎2
4
)
|𝑞1| + |𝑞2| − |𝑞3| − |𝑞4|
2
2
𝐸𝑥 =
1
4𝜋𝜀𝑎2
|𝑞1| + |𝑞2| − |𝑞3| − |𝑞4| √2
𝐸𝑥 =
|𝑞1| + |𝑞2| − |𝑞3| − |𝑞4| √2
4 ∗ (3,14)(8,85 ∗ 10−12)(5 ∗ 10−2)2
𝐸𝑥 =
|𝑞1| + |𝑞2| − |𝑞3| − |𝑞4| √2
4 ∗ (3,14)(8,85 ∗ 10−12)(5 ∗ 10−2)2
𝐸𝑥 =
|10 10−9 | + | − 20 10−9 | − |20 10−9 | − | − 10 10−9 | √2
4 ∗ (3,14)(8,85 ∗ 10−12)(25 ∗ 10−4)
𝐸𝑥 =
10 10−9 + 20 10−9 − 20 10−9 − 10 10−9 √2
4 ∗ (3,14)(8,85 ∗ 10−12)(25 ∗ 10−4)
𝐸𝑥 = 0 𝑁/𝐶
𝑎
𝑎
q1 -q2
q3-q4
𝑎√2
2
𝐸2
𝐸1
𝐸3
𝐸4
𝐸 =
1
4𝜋𝜀
|𝑞1|
𝑑2
 𝑟
𝐸 = 𝐸1 + 𝐸2 − 𝐸3 − 𝐸4
𝐸𝑦 = −𝐸1 + 𝐸2 + 𝐸3 − 𝐸4 𝑠𝑒𝑛45°
𝐸𝑦 =
1
4𝜋𝜀𝑑2
− 𝑞1 + 𝑞2 + |𝑞3| − |𝑞4| 𝑠𝑒𝑛45°
𝐸𝑦 = −
1
4𝜋𝜀
𝑞1
𝑑2
+
1
4𝜋𝜀
𝑞2
𝑑2
+
1
4𝜋𝜀
|𝑞3|
𝑑2
−
1
4𝜋𝜀
|𝑞4|
𝑑2
𝑠𝑒𝑛45°
𝐸𝑦 =
1
4𝜋𝜀(
𝑎 2
2
)2
− 𝑞1 + 𝑞2 + |𝑞3| − |𝑞4| 𝑠𝑒𝑛45°
𝑎
𝑎
q1 -q2
q3-q4
𝑎√2
2
𝐸2
𝐸1
𝐸3
𝐸4
𝐸𝑦 =
1
4𝜋𝜀(
𝑎 2
2
)2
− 𝑞1 + 𝑞2 + |𝑞3| − |𝑞4| 𝑠𝑒𝑛45°
𝐸𝑦 =
1
4𝜋𝜀(
2𝑎2
4
)
− 𝑞1 + 𝑞2 + |𝑞3| − |𝑞4|
2
2
𝐸𝑦 =
1
4𝜋𝜀𝑎2
− 𝑞1 + 𝑞2 + |𝑞3| − |𝑞4| √2
𝐸𝑦 =
− 𝑞1 + 𝑞2 + |𝑞3| − |𝑞4| √2
4 ∗ (3,14)(8,85 ∗ 10−12)(5 ∗ 10−2)2
𝐸𝑦 =
− 𝑞1 + 𝑞2 + |𝑞3| − |𝑞4| √2
4 ∗ (3,14)(8,85 ∗ 10−12)(5 ∗ 10−2)2
𝐸𝑦 =
− 10 10−9 + −20 10−9 + |20 10−9 | − | − 10 10−9 | √2
4 ∗ (3,14)(8,85 ∗ 10−12)(25 ∗ 10−4)
𝐸𝑦 =
−10 10−9 + 20 10−9 + 20 10−9 − 10 10−9 √2
4 ∗ (3,14)(8,85 ∗ 10−12)(25 ∗ 10−4)
𝐸𝑦 =
−10 10−9 + 20 10−9 + 20 10−9 − 10 10−9 √2
4 ∗ (3,14)(8,85 ∗ 10−12)(25 ∗ 10−4)
𝐸𝑦 =
20 10−9 √2
4 ∗ (3,14)(8,85 ∗ 10−12)(25 ∗ 10−4)
𝐸𝑦 =
20 10−9 √2
2789,31 ∗ 10−16
𝐸𝑦 =
28,28 107
2789,31
𝐸𝑦 = 0,012 𝑁/𝐶
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