Question

A charge of -3.40 nC is placed at the origin of an xy-coordinate system, and a charge of 2.45 nC is placed on the y axis at y = 4.25 cm . a.If a third charge, of 5.00 nC , is now placed at the point x = 2.90 cm , y = 4.25 cm find the x and y components of the total force exerted on this charge by the other two charges.
b.Find the magnitude of this force.
c.Find the direction of this force.

Answer 1

Answer:

The magnitude of this force is $1.866\times10^{-4}\ N$

The direction of this force is 55.69°.

Explanation:

Given that,

Charge at origin $q_{1}= -3.40\ nC$

Charge at y axis $q_{2}= 2.45\ nC$

Distance on y axis = 4.25 cm

Third charge $q_{3}= 5.00\ nC$

Distance on x axis = 2.90 cm

(a). We need to calculate the force F₁₃  

Using formula of force

$F_{13}=\dfrac{kq_{1}q_{3}}{r^2}$

Put the value into the formula

$F_{13}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times5.00\times10^{-9}}{(2.90\times10^{-2})^2}$

$F_{13}=-0.00018192\ N$

$F_{13}=-1.82\times10^{-4}\ N$

We need to calculate the force F₁₂  

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{r^2}$

Put the value into the formula

$F_{12}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times2.45\times10^{-9}}{(4.25\times10^{-2})^2}$

$F_{12}=-0.00004150\ N$

$F_{12}=-4.15\times10^{-5}\ N$

The magnitude of this force is

The total force exerted on this charge by the other two charges.

$F=\sqrt{F_{13}^2+F_{12}^2+2F_{13}F_{12}\cos \theta}$

Here, $\theta=90$

$F=\sqrt{F_{13}^2+F_{12}^2}$

Put the value into the formula

$F=\sqrt{(-1.82\times10^{-4})^2+(-4.15\times10^{-5})^2}$

$F=0.0001866\ N$

$F=1.866\times10^{-4}\ N$

(c). We need to calculate the direction of this force

Using formula of direction

$\tan\theta=\dfrac{y}{x}$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{4.25}{2.90})$

$\theta=55.69^{\circ}$

Hence, The magnitude of this force is $1.866\times10^{-4}\ N$

The direction of this force is 55.69°.