This question is based on the electric field. This is the question which I would like to solve using this integral: This is the equation for Electric Field. $$E=\int \frac{dq}{r^s}$$ I was wondering if this approach was valid?
This is volume charge density and an exression can developed that states:
$$dq=\rho\ dV$$ $$dq = \rho 4\pi r^2 \ dr$$ $$\rho = \frac{q_{in}}{V_{in}}=\frac{Q_{total}}{V_{total}}$$
Which leads me to say then that:
$$\frac{q_{in}}{\frac{4}{3}\pi a^3}\cdot\frac{4}{3} \pi r^3=Q$$ Then when I take the derivative of both sides I get:
$$\frac{3q_{in}r^2}{a^3}dr=dQ$$
$$k\int \frac{3q_{in}r^2}{r^2a^3}dr$$ $$E=3k\frac{q_{in}}{a^3}r$$
The formula next to last is not correct. Let's say that the original sphere has radius $a$, and we want to calculate the electric field at some distance $R$ from the center. A spherical shell that you use for integration has a radius $r$, with $0\le r\le R$. The electric field due to this shell is $\frac{dQ}{R^2}$, and not $\frac{dQ}{r^2}$ as you have used in your calculations. Then, you get $$E=k\int_0^R\frac{dQ(r)}{R^2}=k\frac 1{R^2}q_{in}=k\frac{Q}{R^2}\frac{R^3}{a^3}=\frac{kQR}{a^3}$$