Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de G
1 answer:
Answer:
The magnitude of the electric field is 1.124 X 10⁷ N/C
Explanation:
Magnitude of electric field is given as;
where;
E is the magnitude of the electric field, N/C
q is the point charge, C
k is coulomb's constant, Nm²/C²
r is the distance of the point charge, m
Given;
q = 5mC = 5×10⁻³ C
r = 2m
k = 8.99 × 10⁹ Nm²/C²
Substitute these values and solve for magnitude of electric field E
Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C
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Answer:
A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.
Explanation:
At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:
(1)
Where:
- Period of rotation of the satellite, measured in seconds.
- Distance of the satellite with respect to the center of the planet, measured in meters.
- Gravitational constant, measured in newton-square meters per square kilogram.
- Mass of the Earth, measured in kilograms.
Now we clear the distance of the satellite with respect to the center of the planet:
(2)
If we know that ,
and
, then the distance of the satellite is:
The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:
(3)
Where:
- Mass of the satellite, measured in kilograms.
- Force exerted on the satellite by the Earth, measured in newtons.
If we know that ,
,
and
, then the gravitational force is:
A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.
Given:
L = 1 mH = H
total Resistance, R = 11
current at t = 0 s,
= 2.8 A
Formula used:
Solution:
Using the given formula:
current after t = 0.5 ms =
for the inductive circuit:
I =0.011 A