Question

What is the net force exerted by these two charges on a third charge q3 = 49.0nC placed between q1 and q2 at x3 = -1.085m ?Your answer may be positive or negative, depending on the direction of the force.Express your answer numerically in newtons to three significant figures.

Answer 1

Answer:

The net force exerted by these two charges on a third charge is $5.468\times10^{-6}\ N$

Explanation:

Given that,

Third charge $q_{3}=49.0\ nC$

Distance$x_{3}=-1.085\ m$

Suppose The magnitude of the force F between two particles with charges Q and Q' separated by a distance d. Consider two point charges located on the x axis one charge, q₁ = -12.5 nC , is located at x₁ = -1.650 m, the second charge, q₂ = 31.5 nC , is at the origin.

We need to calculate the total force will be the vector sum of two forces

Using Coulomb's law,

$F_{13}=\dfrac{kq_{1}q_{3}}{(x_{1}-x_{3})^2}$

Put the value into the formula

$F_{13}=\dfrac{9\times10^{9}\times(-12.5\times10^{-9})\times49\times10^{-9}}{(-1.650-(-1.085))^2}$

$F_{13}=-17268.3\times10^{-9}\ N$

We need to calculate the force will be to the negative charge with opposite charges

Using Coulomb's law,

$F_{23}=\dfrac{kq_{2}q_{3}}{(x_{2}-x_{3})^2}$

Put the value into the formula

$F_{23}=\dfrac{9\times10^{9}\times(31.5\times10^{-9})\times49\times10^{-9}}{(-1.085)^2}$

$F_{23}=11800.2\times10^{-9}\ N$

The force also will be to the negative side, charges with same charge sign

We need to calculate the net force exerted by these two charges on a third charge

Using formula of net force

$F_{net}=F_{13}+F_{23}$

$F_{net}=-17268.3\times10^{-9}+11800.2\times10^{-9}$

$F_{net}=-0.0000054681\ N$

$F_{net}=-5.468\times10^{-6}\ N$

Negative sign shows the negative direction.

Hence, The net force exerted by these two charges on a third charge is $5.468\times10^{-6}\ N$