The magnitude of the test charge must be small enough so that it does not disturb the issuance of the charges whose electric field we wish to measure otherwise the metric field will be different from the actual field.
<h3>How does test charge affect electric field?</h3>
As the quantity of authority on the test charge (q) is increased, the force exerted on it is improved by the same factor. Thus, the ratio of force per charge (F / q) stays the same.
Adjusting the amount of charge on the test charge will not change the electric field force.
<h3>What is a test charge used for?</h3>
The charge that is used to measure the electric field strength is directed to as a test charge since it is used to test the field strength. The test charge has a portion of charge denoted by the symbol q.
To learn more about test charge, refer
brainly.com/question/16737526
#SPJ9
Answer:
225 N
Explanation:
"Below the horizontal" means he's pushing down at an angle.
Draw a free body diagram of the box. There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.
Sum of forces in the y direction:
∑F = ma
N − mg − F sin θ = 0
N = F sin θ + mg
Plug in values:
N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)
N = 225 N
The wavelength is 2m.
Hence, Option c) 2m is the correct answer
Given that;
Frequency;
Speed; 
Wavelength; 
using the expression for the relations between wavelength, frequency and speed of wave:

Where
is wavelength, f is frequency and v is speed.
We substitute our given values into the equation

The wavelength is 2m.
Hence, Option c) 2m is the correct answer.
To learn more about wavelength, click here: brainly.com/question/1347107
Answer:
To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is
if the resistance > 1( very high ) then component is a capacitor
if the resistance = 0 then component is an inductor
Explanation:
To determine the mystery component we will connect the mystery component to a DC voltage source, then I will measure the resistance of the component with the use of Ohmmeter, the value of the resistance of the mystery component will determine what the mystery component is
if the resistance > 1( very high ) then component is a capacitor
if the resistance = 0 then component is an inductor