Question

Three charges form a right triangle. A charge q1= -3 μC is 4 cm above a charge q2= 6 μC. A charge q3= 2 μC is 3 cm to the right of the chargeq2. What is the magnitude and direction of the resultant force on the charge2 q?

Answer 1

Given that there are three charges forming the right triangle such that

`q1\text{ = -3 }\mu C=-3*10^{-6\text{ }}C`

is placed 4 cm = 0.04 m from the charge

`q2=\text{ 6 }\mu C=6*10^{-6\text{ }}C`

Another charge

`q3=\text{ 2}\mu C=2*10^(-6)\text{ C}`

is placed 3 cm = 0.03 m right from the charge q3.

Here, Coulomb's law can be used to calculate force,

`F=k(q1q2)/(r^2)`

Here, F is the force, q1 and q2 are the charges, r is the distance between q1 and q2 and k is a constant whose value is

`9*10^9Nm^2/C^2`

Substituting values for the force on q2 due to q1, we get

`\begin{gathered} F_(12)=(9*10^9*(3*10^(-6))*6*10^(-6))/((0.04)^2) \\ =\text{ }1.01\text{ N} \end{gathered}`

Substituting values for the force on q2 due to q3, we get

`\begin{gathered} F_(32)=(9*10^9*(2*10^(-6))*6*10^(-6))/((0.03)^2) \\ =\text{ 120 N} \end{gathered}`

Hence the total force is

`\begin{gathered} F_2=F_(12)+F_(32) \\ =120\text{ +1.01} \\ =\text{ }121.01\text{ N} \end{gathered}`

Here, A is the attractive force due to q1 and B is the repulsive force due to q3

R is the resultant due to both of them.

R=A+B

Please note there are arrows above them.

Thus R is the resultant direction.