Answer:
Explanation:
The relationship between the current and the electric field inside a wire is given by the following equation:
where ρ is the resistivity, and J is the current density. The current density is the current per area:
The resistivity is the reciprocal of conductivity. Therefore,
Finally, the electric field can be found:
Everything in the universe that's visible, except stars,
is visible because it reflects some of the light that
shines on it.
Based on the values of E, C, and R, the maximum charge on the capacitor can be found to be 1.80 x 10⁻⁴ C.
<h3>What is the maximum charge?</h3>
The maximum charge on the capacitor can be found as:
= C x ( 10⁻⁶ / 1 μF) x E
Solving gives:
= 20.0 x ( 10⁻⁶ / 1 μF) x 9.00
= 1.80 x 10⁻⁴ C
In conclusion, the maximum charge is 1.80 x 10⁻⁴ C
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I think you want to determine the exit speed?
You have to determine how much velocity was decreased by calculating it from the kinetic energy.
KE = (1/2)mv²
1.4 x 10^5 = (1/2)*(1100)v²
v² = 254.55
v =15.95 m/s
So the velocity reduces by 15.95 m/s. Subtracting this to the initial velocity: 22 - 15.95 = 6.05 m/s.
So, the final speed was 6.05 m/s.
I hope I was able to help :)
