Question

two equal positive charges each of 2 microcoulomb interact with third positive charge as shown in figure find the magnitude and direction of force experienced by the charge 3 microcoulomb ​

Answer 1

solutions is attached above....... .

Answer 2

Answer:

the magnitude of force experienced by the charge 3 microcoulomb ​is 0.003456 N and along the direction "OC".

Explanation:

Given, Two charges of $2\mu C$ are placed at "A" and "B" corners of the triangle

and the third charge of $3\mu C$ is placed at corner "C".

Here, we define angle∠ACO = ∠OCB = θ

From the hypothe we get $AC = BC = \sqrt{3^2+4^2} = 5$

Now the force acting along the side "AC" is ($F_{AC} = \frac{k(2\mu C)(3\mu C)}{AC^2}$)...(1).

We try to resolve the force $F_{AC}$ along the horizontal direction and vertical direction.

i.e we get          

$F_{AC}^{hori}= F_{AC}\cos(\theta)\\ F_{AC}^{verti} = -F_{AC}\sin(\theta)$....(2)  (The vertical component of force along AC is negative. Since, the force is acting downwards)

We substitute equation (1) in equation (2)

$F_{AC}^{hori}=\frac{k(2\mu C)(3\mu C)}{AC^2}\cos(\theta)\\ F_{AC}^{verti} = -\frac{k(2\mu C)(3\mu C)}{AC^2}\sin(\theta)$

From Δ AOC we get cos(θ) = 4/5 and sin(θ) = 3/5, and also AC = 5

Therefore,  

  $F_{AC}^{hori}=\frac{k(2\mu C)(3\mu C)\times 4}{5^2\times 5}\\ F_{AC}^{verti} = -\frac{k(2\mu C)(3\mu C)\times 3}{5^2\times 5}$

$F_{AC}^{hori}=\frac{9\times 10^9(2\times10^{-6} C)(3\times10^{-6}C)\times 4}{5^2\times 5}\\ F_{AC}^{verti} = -\frac{9\times10^9(2\times10^{-6}C)(3\times10^{-6} C)\times 3}{5^2\times 5}$

$F_{AC}^{hori}=\frac{0.216}{125}\\ F_{AC}^{verti} = -\frac{0.162}{125}$

Therefore,

$F_{AC}^{hori}=0.001728\\ F_{AC}^{verti} =-0.001296$  

And the force acting along the side "BC" is ($F_{BC} = \frac{k(2\mu C)(3\mu C)}{BC^2}$)....(3)

We try to resolve the force $F_{BC}$ along the horizontal direction and vertical direction.

i.e. we get

$F_{BC}^{hori}= F_{BC}\cos(\theta)\\ F_{BC}^{verti} = F_{BC}\sin(\theta)$...(4)

We substitute equation (3) in equation (4)

$F_{BC}^{hori}=\frac{k(2\mu C)(3\mu C)}{AC^2}\cos(\theta)\\ F_{BC}^{verti} = \frac{k(2\mu C)(3\mu C)}{AC^2}\sin(\theta)$

From Δ BOC we get cos(θ) = 4/5 and sin(θ) = 3/5, and also AC = 5

Therefore,  

  $F_{BC}^{hori}=\frac{k(2\mu C)(3\mu C)\times 4}{5^2\times 5}\\ F_{BC}^{verti} = \frac{k(2\mu C)(3\mu C)\times 3}{5^2\times 5}$

$F_{BC}^{hori}=\frac{9\times 10^9(2\times10^{-6} C)(3\times10^{-6}C)\times 4}{5^2\times 5}\\ F_{BC}^{verti} = \frac{9\times10^9(2\times10^{-6}C)(3\times10^{-6} C)\times 3}{5^2\times 5}$

$F_{BC}^{hori}=\frac{0.216}{125}\\ F_{BC}^{verti} = \frac{0.162}{125}$

Therefore,

$F_{BC}^{hori}=0.001728\\ F_{BC}^{verti} =0.001296$.

Now to obtain the final result we add up the forces along the vertical and horizontal direction.

i.e. both the horizontal forces add up whereas both the vertical forces gets nullified.

Therefore, $F_{res}= F_{AC}^{hori}+F_{BC}^{hori} = 0.003456 N$ acting along the direction "OC".

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