Solve Electromagnetism: Gold Nugget Net Charge Problem

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To solve the gold nugget net charge problem, first determine the mass of a gold atom and calculate the total number of gold atoms in a 1.0-gram nugget. Next, calculate the total number of electrons by multiplying the number of gold atoms by the atomic number of gold. After removing 1.0% of the electrons, the net charge can be found by considering the remaining electrons and the protons present. The discussion also briefly touches on calculating the electric field inside and outside a uniformly polarized cylinder, indicating a mathematical relationship for the external field.
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1. Suppose a 1.0-g nugget of pure gold has zero net charge. What would be its net charge after it has 1.0% of its electrons removed?
please help! i don't know what to do
 
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1. Find the mass of a gold atom from a table.
2. Obtain the number of gold atoms. i.e. (# of atoms) = 1.0g / (mass of a gold atom)
3. (# of electrons) = (# of gold atoms) x (atomic number of gold)

I guess this is enough for the hint?
 
after obtaining the electrons, i will get the 1% of electrons removed then ill add the electrons and protons to obtain the net charge? am i correct?
 
A very long cylinder of radius R, carries a uniform polarisationP perpendicular to it's axis.Find the electric field in side the cylinder. Show that the field outside the cylinder can be expressed in the form
E=Rsquared divided
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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