The strength of an electric field depends on the
2 answers:
<span>The correct option is D) both A and B. In fact, the strength of an electric field produced by a single point charge is given by
</span>
<span>
where Q is the charge that generates the field, k is the Coulomb's constante and r is the distance from the charge source of the field.
From the equation, we see that the strength of the field depends on A) the amount of charge that produced the field (Q) and B) the distance from the charge (r), so the correct option is D.</span>
</span>
where Q is the charge that generates the field, k is the Coulomb's constante and r is the distance from the charge source of the field.
From the equation, we see that the strength of the field depends on A) the amount of charge that produced the field (Q) and B) the distance from the charge (r), so the correct option is D.</span>
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Answer:
Explanation:
According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
Here k is the Coulomb constant. In this case, we have ,
and
. Replacing the values:
The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:
Answer:
the question is incomplete, the complete question is
"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"
2.6mA
Explanation:
we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.
using the formula be low,
where B is the magnitude of the field and A is the area of the circular coil.
First, let determine the area using where r is the radius of 5cm or 0.05m
since we no that the angle is at
we determine the magnitude of the magnetic filed
the Magnitude of the voltage is 0.000532V
Next we determine the current using ohm's law