The strength of the electric field 0.5 m from a 6 µc charge is n/c. (use k = 8.99 × 109 n•meters squared per coulomb squared and
1 answer:
The strength of the electric field will be 215760 NC⁻¹. The concept of the electric field strength is used in the problem.
<h3>What is electric file strength?</h3>The electric field strength is defined as the ratio of electric force and charge. Its unit is NC⁻¹.
The given data in the problem is;
E is the Electric Field Strength
k is the Colomb's constant = 8.99 x 10⁹ N.m²/C²
q is the magnitude of charge = 6 μC = 6 x 10⁻⁶ C
r is the distance = 0.5 m
The electric field strength is given by the formula;
Hence the strength of the electric field will be 215760 NC⁻¹.
To learn more about the electric field strength refer to the link;
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Answer:
Explanation:
Our Given Parameters include:
Average radius (r) for each parallel coil = 12.5 cm = 12.5 × 10⁻² m
The lower coil turns = 20 turns
The lower coil constant current = 1.30 A
The upper is suspended 0.314 cm above the lower coil
That indicates the distance = 0.314 cm = 0.314 × 10⁻² m
Current for the upper coil = ???(unknown)
Force (F) = 3.30 N
If we take each coil into cognizance as a long parallel straight wire; then the length of the lower coil can be calculated as:
where:
= number of turns of the lower coil = 20 turns
r = average radius in the lower coil = 12.5 × 10⁻² m
Substituting our values; we have:
From our parameters above:
= constant current in the lower coil = 4.0 A
But the magnitude of the magnetic force (F) = 1 LB
Then the force on the lower coil in regard to the upper coil can be :
F =
Making the varied current in the upper coil the subject of the formula; we have:
where :
Then:
≅ 2539 A
However; the current needed in the upper coil in each turn will be:
Answer:
The value of variable capacitor is F
Explanation:
Given :
Inductance H
Frequency Hz
According to the cutoff frequency,
Now we find the value of capacitance,
F
Therefore, the value of variable capacitor is F