Question

Consider three charges located in the xy plane. A charge of +3 μC is located at x=4.00 cm and y = 0, a charge of – 2.00 μC is located at x=0 and y= 5.00cm. Findthe netforce on a charge of + 6 μC located at the origin.

Answer 1

Answer:

The vector expression of net force on charge 3 is

$\vec{F}_3 = -101.7~\^x + 43.4~\^y$

The magnitude of the net force is 110 N.

Explanation:

We will use Coulomb's Law to find the forces.

$\vec{F} = \frac{1}{4\pi \epsilon_0}\frac{q_1q_2}{r^2}\^r$

In this case, there are three charges and we need to find the forces on the charge 3 exerted by charges 1 and 2 separately.

For the force that charge 1 exerts on charge 3:

$\vec{F}_{1on3} = \frac{1}{4\pi \epsilon_0}\frac{(3\times 10^{-6})(6\times 10^{-6})}{(4\times 10^{-2})^2}(-\^x) = -101.7~\^x$

The force is in the -x direction, because they are both positive, so they repel each other.

For the force that charge 2 exerts on charge 3:

$\vec{F}_{2on3} = \frac{1}{4\pi \epsilon_0}\frac{(2\times 10^{-6})(6\times 10^{-6})}{(5\times 10^{-2})}(\^y) = 43.4~\^y$

The force is in the +y direction, because they have opposite charges, so they attract each other.

The net force on charge 3 is

$\vec{F}_3 = -101.7~\^x + 43.4~\^y$

The magnitude of the net force is 110 N.