Question

Two charged particles are a distance of 1.82 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.14 nC.

Answer 1

The magnitude of the electric force is(F) $79.07\times 10^{-9}$ N and the electric force between the charges is repulsive.

How can we calculate the magnitude of the electric force?

To calculate the magnitude of the electric force that one particle exerts on the other,  we are using the formula,

$F=k\frac{q_{1} q_{2} }{r^{2} }$

Here we are given,

$q_1$= Charge of the first particle= 7.03nC = $7.03\times 10^{-9}$  C.

$q_2$=Charge of the second particle = 4.14nC= $4.14\times 10^{-9}$ C.

r= The distance between two charges=1.82m.

k= Constant of proportionality= $\frac{1}{4\pi \epsilon}$=$9\times 10^{9} Nm^{2} /C^{2}$

Now we put the known values in the above equation, we get

$F=k\frac{q_{1} q_{2} }{r^{2} }$

Or,$F=9\times 10^{9} \times \frac{7.03\times 10^{-9}\times 4.14\times 10^{-9} }{(1.82)^{2} }$

Or, F= $79.07\times 10^{-9}$ N.

From the above calculation we can conclude that the magnitude of the electric force(F) is $79.07\times 10^{-9}$ N.

Charges $q_1$ and $q_2$ is $79.07\times 10^{-9}$ C and $4.14\times 10^{-9}$C respectively, Both charges are positive and have same polarity. The force between the same charges is repulsive.

Thus, we can conclude that, the electric force between the charges is repulsive.

Learn more about Charged particle:

brainly.com/question/18763828

#SPJ4

Disclaimer: The question was given incomplete in the portal. Here is the complete question.

Question: Two charged particles are a distance of 1.82 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.14 nC.

• What is the magnitude of the electric force that one particle exerts on the other?

• Is the force attractive or repulsive?