Question

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.0 cm. What is the magnitude of the electric force on a third charge (+2.5 µC) placed on the y axis at y = 3.0 cm?

Answer 1

Answer:

The magnitude of electric force is  $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

• directly proportional to the products of charges i.e $F\propto q_1q_2$

• inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$    [ 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, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

                         $=2k\frac{Qq_1}{r^2} cos\theta \hat i$

                         $=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

                         $=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$     [ r=5]

                         $=7.2\times10^{-3}\hat i$   N

The magnitude of electric force is  $7.2\times10^{-3} N$