Name: ____________________________
Physics I C
Date: _____________
Mr. Tiesler
Solutions to Electric Fields & Forces Problems 6-10
6.) A strong lightning bolt transfers about 25 C to Earth. How many electrons are transferred?
𝑞 = 25 𝐶
−25 𝐶 (
1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
) = 1.6𝑥1020 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
−1.60𝑥10−19 𝐶
7.) Two electrons in an atom are separated by 1.5x10-10 m, the typical size of an atom. What is
the electric force between them?
𝑟 = 1.5𝑥10−10 𝑚
𝑞 = −1.60𝑥10−19 𝐶
𝑘 = 9.0𝑥109 𝑁 ∙ 𝑚2 ⁄𝐶 2
𝐹 =?
|𝑞1 𝑞2 |
|𝑞 2 |
|(−1.60𝑥10−19 𝐶)2 |
9
2
2
𝐹=𝑘
= 𝑘 2 = (9.0𝑥10 𝑁 ∙ 𝑚 ⁄𝐶 )
= 1.0𝑥10−8 𝑁, 𝑎𝑤𝑎𝑦
(1.5𝑥10−10 𝑚)2
𝑟2
𝑟
8.) Two identical positive charges exert a repulsive force of 6.4x10-9 N when separated by a
distance of 3.8x10-10 m. Calculate the charge of each.
𝐹 = 6.4𝑥10−9 𝑁
𝑟 = 3.8𝑥10−10 𝑚
𝑘 = 9.0𝑥109 𝑁 ∙ 𝑚2 ⁄𝐶 2
𝑞 =?
|𝑞1 𝑞2 |
|𝑞 2 |
𝐹=𝑘
=𝑘 2 ,
𝑟2
𝑟
(6.4𝑥10−9 𝑁)(3.8𝑥10−10 𝑚)2
𝐹𝑟 2
√
√
𝑞=
=
= 3.2𝑥10−19 𝐶
9
2
2
⁄
𝑘
9.0𝑥10 𝑁 ∙ 𝑚 𝐶
9.) A positive charge of 3.0 µC is pulled on by two negative charges. As shown in Figure 2, one
negative charge, -2.0 µC, is 0.050 m to the west, and the other, -4.0 µC, is 0.030 m to the east.
What total force is exerted on the positive charge?
𝑞1 = 3.0 𝜇𝐶 = 3.0𝑥10−6 𝐶
𝑞2 = −2.0 𝜇𝐶 = −2.0𝑥10−6 𝐶
𝑞3 = −4.0 𝜇𝐶 = −4.0𝑥10−6 𝐶
Figure 2
𝑟12 = 0.050 𝑚
𝑟13 = 0.030 𝑚
𝑘 = 9.0𝑥109 𝑁 ∙ 𝑚2 ⁄𝐶 2
𝐹𝑛𝑒𝑡 =?
𝐹12 = 𝑘
|𝑞1 𝑞2 |
|(3.0𝑥10−6 𝐶)(−2.0𝑥10−6 𝐶)|
9
2⁄ 2)
(9.0𝑥10
=
𝑁
∙
𝑚
𝐶
= 22 𝑁
(0.050 𝑚)2
𝑟12 2
Because charges 1 and 2 have opposite charges, the force exerted on charge 1 by charge 2 is in
the negative-x direction.
𝐹13 = 𝑘
|𝑞1 𝑞3 |
|(3.0𝑥10−6 𝐶)(−4.0𝑥10−6 𝐶)|
9
2⁄ 2)
(9.0𝑥10
=
𝑁
∙
𝑚
𝐶
= 120 𝑁
(0.030 𝑚)2
𝑟13 2
Because charges 1 and 3 have opposite charges, the force exerted on charge 1 by charge 3 is in
the positive-x direction.
𝐹𝑛𝑒𝑡 = 120 𝑁 − 22 𝑁 = 98 𝑁
10.) Three particles are placed in a line. The left particle has a charge of -55 µC, the middle one
has a charge of +45 µC, and the right one has a charge of -78 µC. The middle particle is 72 cm
from each of the others, as shown in Figure 3. (a) Find the net force on the middle particle.
𝑞1 = −55 𝜇𝐶 = −55𝑥10−6 𝐶
𝑞2 = 45 𝜇𝐶 = 45𝑥10−6 𝐶
𝑞3 = −78 𝜇𝐶 = −78𝑥10−6 𝐶
Figure 3
𝑟12 = 72 𝑐𝑚 = 0.72 𝑚
𝑟23 = 72 𝑐𝑚 = 0.72 𝑚
𝑘 = 9.0𝑥109 𝑁 ∙ 𝑚2 ⁄𝐶 2
𝐹𝑛𝑒𝑡 =?
𝐹21 = 𝑘
|𝑞1 𝑞2 |
|(−55𝑥10−6 𝐶)(45𝑥10−6 𝐶)|
9
2⁄ 2)
(9.0𝑥10
=
𝑁
∙
𝑚
𝐶
= 43 𝑁
(0.72 𝑚)2
𝑟12 2
Because charges 1 and 2 have opposite charges, the force exerted on charge 2 by charge 1 is in
the negative-x direction.
𝐹23 = 𝑘
|𝑞2 𝑞3 |
|(45𝑥10−6 𝐶)(−78𝑥10−6 𝐶)|
9
2⁄ 2)
(9.0𝑥10
=
𝑁
∙
𝑚
𝐶
= 61 𝑁
(0.72 𝑚)2
𝑟23 2
Because charges 2 and 3 have opposite charges, the force exerted on charge 2 by charge 3 is in
the positive-x direction.
𝐹𝑛𝑒𝑡 = 61 𝑁 − 43 𝑁 = 18 𝑁
(b) Find the net force on the right particle.
𝐹31
|𝑞1 𝑞3 |
|(−55𝑥10−6 𝐶)(−78𝑥10−6 𝐶)|
9
2⁄ 2)
(9.0𝑥10
=𝑘
=
𝑁∙𝑚 𝐶
= 18.6 𝑁
(1.44 𝑚)2
𝑟13 2
Because charges 1 and 3 have the same charge, the force exerted on charge 3 by charge 1 is in
the positive-x direction.
𝐹23
|𝑞2 𝑞3 |
|(45𝑥10−6 𝐶)(−78𝑥10−6 𝐶)|
9
2
2
=𝑘
= (9.0𝑥10 𝑁 ∙ 𝑚 ⁄𝐶 )
= 61 𝑁
(0.72 𝑚)2
𝑟23 2
Because charges 2 and 3 have opposite charges, the force exerted on charge 3 by charge 2 is in
the positive-x direction.
𝐹𝑛𝑒𝑡 = 18.6 𝑁 + 61 𝑁 = 79.6 𝑁 = 8.0𝑥101 𝑁