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.
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Answer:
Angle of reflection of light is 34 degree
Explanation:
As per law of reflection of light we know that
angle of incidence of light = angle of reflection of light
So here we know that
angle of incidence on the surface of oil is given as
so we know that
so here we can say that reflection angle of light will be same as angle of incidence
Answer:
Explanation:
a ) Thermal efficiency = work output / heat input
= .38 MW / 1 MW = .38
OR 38%
Heat rejected at cold reservoir = heat input - work output
1 MW - .38 MW
= 0.62 MW.
b ) For reversible power output
efficiency = T₂ - T₁ / T₂ ; T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.
= 1200 - 300 / 1200 = 900 / 1200
= .75
or 75%
rate at which heat is rejected
= 1 - .75 x 1
= .25 MW .
Answer:20°
Explanation:
Recall
Range R of a projectile is given by U^2sin2A/g
We're U = velocity,A= angle of projection and g is acceleration due to gravity
From the question the range R are the same
Hence R1=R2
U1^2sin2A/g=U2^2sin2B/g
But U1=U2 and g=g
Hence sin2A=sin 2B
Sin 2*70= sin2*B
0.6427=sin2B
B=sin inverse(0.6427)=40/2=20°
