Respuesta :
Answer:
The magnitude of electric force is [tex]7.2\times10^{-3} N[/tex]
Explanation:
Coulomb's Law:
The force of attraction or repletion is
- directly proportional to the products of charges i.e [tex]F\propto q_1q_2[/tex]
- inversely proportional to the square of distance i.e [tex]F\propto \frac{1}{r^2}[/tex]
[tex]\therefore F\propto \frac{q_1q_2}{r^2}[/tex]
[tex]\Rightarrow F=k \frac{q_1q_2}{r^2}[/tex] [ k is proportional constant=9×10⁹N m²/C²]
There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C
Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C
and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C
Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.
If we draw a line from q₁ to Q .
The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.
Let hypotenuse = r
Therefore, [tex]r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5[/tex]
we know,
[tex]cos \theta = \frac{base }{hypotenuse}[/tex]
[tex]\Rightarrow cos \theta = \frac{4 }{r}[/tex]
Total force [tex]F_Q = 2.F_1 cos\theta \hat{i}[/tex]
[tex]=2k\frac{Qq_1}{r^2} cos\theta \hat i[/tex]
[tex]=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i[/tex]
[tex]=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i[/tex] [ r=5]
[tex]=7.2\times10^{-3}\hat i[/tex] N
The magnitude of electric force is [tex]7.2\times10^{-3} N[/tex]
