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
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Answer:
3.25 × 10^7 m/s
Explanation:
Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)
Ek = 1/2 mv2 = qΔV .................. 1
Given that V is the electron speed in m/s
Charge of electron = 1.60217662 × 10-19 coulombs
Mass of electron = 9.109×10−31 kilograms
ΔV = 3.0kV = 3000V
Make V the subject of the formula in eqaution 1
V = sqr root 2qΔV/m
V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31
V = 3.25 × 10^7 m/s
I think you almost got it.
At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.
With the value of the initial speed (28 m/s, which is the total speed), you can set
v_x = v * cos( theta ) ---> 12 = 28*cos(theta) --> cos(theta)=12/28=3/7
or theta = 64.62 deg, it is D. Think about it. I hope you see it.
The answer to this question is false
Answer:
<em>Answer: Work equals force times distance. 3,000 J</em>
Explanation:
Work Done By A Force
When some force 
is applied and a displacement 
is achieved, the work done by the force is given by
Note that the work is a scalar magnitude as the result of the dot-product of two vectors. If the force and the displacement are parallel, then the vectors can be replaced as its magnitudes F,x and the work is
The dot product becomes a simple arithmetic product, i.e force times distance.
Sara weighs 500 Nw and she climbs up a 6 meter set of stairs. She needs to lift her weight up, so the force is the weight and the distance is the height of the stairs, thus
Answer: Work equals force times distance. 3,000 J
