Question

Suppose a charge q=+50 uC experiences a force of 0.040 N which points to the right. What is the magnitude and direction of the electric field that causes this force?

Answer 1

The magnitude of the electric field is 800 N/C and the direction is positive and points to the right.

Step-by-step explanation:

Ok, here is the explanation:

The magnitude of the electric field is simply defined as the force per charge on the test charge:

`electric field strength=(force)/(charge)`

If the electric field strength is denoted by the symbol E, then the equation can be rewritten in symbolic form as:

`E=(F)/(q)`

and this is the equation we are going to use to solve the problem

We have: E=?, q=+50μC and F=0.040N

We know that 1C=1x10-6 μC so 50μC is equal to 5x10-5 C or 0.00005 C

Then we use the equation as follows:

`E=(0.040)/(+0.00005) =800 (N)/(C)`

Because the charge is positive and also the force applied, so will the electric field. And since the force points to the right, the electric field as a vector points to the right.