<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
Answer:
A-the energy of the wave decreases gradually
Explanation:
when a wave is acted upon by an external damping force the energy of the wave decreases gradually.
The energy degrades into the form of heat which is considered to be of less value and use. The reason is because it disperses and spreads more widely.
So therefore it end up as heat with a little sound but that is close to none because that too disperses into heat i.e. decreased form of energy.
| Impedance | = √ [R² +(ωL)²]
R² = 6800² = 4.624 x 10⁷
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²
= 2.0884 x 10⁻⁴ f²
| Z | = √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ] = 1.6 x 10⁵
(1.6 x 10⁵)² = (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)
(2.56 x 10¹⁰) - (4.624 x 10⁷) = 2.0884 x 10⁻⁴ f²
Frequency² = (2.56 x 10¹⁰ - 4.624 x 10⁷) / 2.0884 x 10⁻⁴
= 2.555 x 10¹⁰ / 2.0884 x 10⁻⁴
= 1.224 x 10¹⁴
= 122,400 GHz <== my calculation
11.1 MHz <== online impedance calculator
Obviously, I must have picked up some rounding errors
in the course of my calculation.
