If a segment of wire has a net charge (uniformly distributed), and then its length is increased by 50%,by what factor has its charge density changed?
If the original charge was 10.0 nC, how much charge must then be added to restore the wire’s original charge
density?
double the length, double the surface area, so charge density is halved.
add another 10nC to bring the sigma back up to original.
To calculate how the charge density of the wire has changed, we need to use the formula for charge density, which is given by:
Charge density (ρ) = Charge (Q) / Length (L)
Let's call the original length of the wire "L" and the original charge density "ρ₁". After increasing the length of the wire by 50%, the new length becomes 1.5L.
To find the new charge density (ρ₂), we divide the original charge (Q) by the new length (1.5L):
ρ₂ = Q / 1.5L
We can simplify this equation by multiplying both the numerator and denominator by 2/3:
ρ₂ = (Q * 2/3) / (1.5L * 2/3)
= (2Q / 3) / (3L / 2)
= (2Q / 3) * (2/3L)
= (4Q / 9L)
Therefore, the charge density has changed by a factor of 4/9 (or 0.4444) when the wire's length is increased by 50%.
To calculate how much charge must be added to restore the wire's original charge density, we need to calculate the difference between the original charge and the charge corresponding to the original charge density.
Let's call the required charge to restore the original charge density "ΔQ".
ΔQ = Original charge (10.0 nC) - Charge corresponding to original charge density (ρ₁ * L)
Substituting the values:
ΔQ = 10.0 nC - (ρ₁ * L)
We can now plug in the original charge density (ρ₁) and the original length (L) to calculate ΔQ.