Answer:
The magnitude of the electric field is 800 N/C and the direction is positive and points to the right.
Explanation:
Ok, here is the explanation:
The magnitude of the electric field is simply defined as the force per charge on the test charge:
[tex]electric field strength=\frac{force}{charge}[/tex]
If the electric field strength is denoted by the symbol E, then the equation can be rewritten in symbolic form as:
[tex]E=\frac{F}{q}[/tex]
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:
[tex]E=\frac{0.040}{+0.00005} =800 \frac{N}{C}[/tex]
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.
