Question

An electric field of ${{10}^{5}}\dfrac{N}{C}$ points due west at a certain spot. What is the magnitude and direction of the force that acts on a charge of $+2\mu c$ and $-5\mu c$at this spot?

Answer 1

To solve the above question where the electric field at a certain point and the electric charges acting on the body are given to us and we have to find the magnitude and direction of the force acting on the point we will use the formula of electric force with respect to electric charge and electric field.

Complete step-by-step solution:
In the above question we have to find the direction and magnitude of the force that is acting on $+2\mu c$and $-5\mu c$if the electric field is given to us which is in due west direction
So, we will first discuss what is force and what is the relationship between the force and the charge.
We know that the force which is due to the electric field on a charge is built into it. It always acts either parallel or antiparallel direction to the electric field and is always independent of the velocity of the charge. This means the force has the ability to do work and impart energy to the charge.
The electric force can be defined as the product of charge acting on the body and electric field at the given spot. Mathematically it can be written as
$F=q.\overrightarrow{E}$
Where, $F$is the electric force, $q$ is the charge acting and $\overrightarrow{E}$is the electric field
Now, we will find the force acting on $+2\mu c$charge
We know that $F=q.\overrightarrow{E}$
Here we have to find the value of $F$that is the electric force, $q$ is$+2\mu c$which is given in the question
But $+2\mu c$= $+2\times {{10}^{-6}}c$
and $\overrightarrow{E}$is ${{10}^{5}}\dfrac{N}{C}$
Putting the values in the formula:
$F=2\times {{10}^{-6}}\times {{10}^{5}}$
$F=0.2N$in the west direction.
Now, we will find the force acting on $-5\mu c$charge
We know that $F=q.\overrightarrow{E}$
Here we have to find the value of $F$that is the electric force, $q$ is $-5\mu c$which is given in the question
But $-5\mu c$ $=-5\times {{10}^{-6}}$
and $\overrightarrow{E}$is ${{10}^{5}}\dfrac{N}{C}$
Putting the values in the formula:

& F=-5\times {{10}^{-6}}\times {{10}^{5}} \\\ & F=-0.5N \\\ \end{aligned}$$ The direction is east due to negative sign and force is $$0.5N$$ **Note:** In questions like these where we have different quantities with different units, always remember to convert all the given quantities into the standard form so that we can reduce the confusion we will have, in the end while getting the answer to the given question.