Why can current be larger than placed charges in a wire?

Click For Summary
Current can exceed the amount of charge placed in a wire because it measures the flow of charge per second, rather than the total charge present. The volume of the wire allows for many small charges to move simultaneously, contributing to a higher current. A larger wire volume reduces resistance, facilitating greater current flow as charges have more space to navigate without colliding with static atoms. The relationship between current (I), charge (Q), and time (t) illustrates this, as a significant amount of charge can pass through in a short time, resulting in a high current. Thus, the apparent discrepancy between large currents and limited charge is explained by the dynamics of charge flow in conductive materials.
plasma tail
Messages
1
Reaction score
0

Homework Statement



We have seen that a coulomb is an enormous amount of
charge; it is virtually impossible to place a charge of 1 C on an
object. Yet, a current of 10 A, is quite reasonable. Explain
this apparent discrepancy.

Homework Equations


I = nqvA

The Attempt at a Solution


i think it is because current is the amount of charge flow per second, and the amount of charge depends on the volume of the wire, hence current can be larger than the placed charges if volume of the wire is pretty big. is this a good answer? is there anything to add? thx
 
Physics news on Phys.org
I agree with: As you've already said, current is defined as the amount of charge flow per second, there could be many charges little charges but summing them up will result in 1 C, bigger volume mean less resistance so more current flow this can be explained (the bad way) because charges have more space to avoid hitting static atoms in the wire (also the reason you've mentionned) and I should mention all the charges flow on the surface,any way if we get into the math, ## I = \frac{\delta Q}{\delta t} ##, if we take for example ##\delta t = 0.01 s## mesure that ##\delta Q = 0.1C## passed, then we conclude that I = 10A, Cheers :D
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...
Thread 'Conducting Sphere and Dipole Problem'
Hi, I'm stuck at this question, please help. Attempt to the Conducting Sphere and Dipole Problem (a) Electric Field and Potential at O due to Induced Charges $$V_O = 0$$ This potential is the sum of the potentials due to the real charges (##+q, -q##) and the induced charges on the sphere. $$V_O = V_{\text{real}} + V_{\text{induced}} = 0$$ - Electric Field at O, ##\vec{E}_O##: Since point O is inside a conductor in electrostatic equilibrium, the electric field there must be zero...