(a) Electric intensity E outside the balloon (r > R)
Using Gauss's law, we can determine the electric intensity E at a distance r from the center of the balloon. We consider a spherical Gaussian surface of radius r, concentric with the balloon. The electric field is everywhere perpendicular to the surface, and its magnitude is constant on the surface. Therefore, the electric flux through the surface is given by:
∮_S \(\overrightarrow E\cdot d\overrightarrow A\)=E⋅4πr^2
The total charge enclosed by the surface is q. Therefore, according to Gauss's law, we have:
∮_S \(\overrightarrow E\cdot d\overrightarrow A\)=\frac{q_{in}}{\varepsilon_0}
where ε₀ is the permittivity of free space. Combining the above equations, we get:
$$E⋅4πr^2=\frac{q}{\varepsilon_0}$$
$$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$$
This is the expression for the electric intensity outside the balloon. It varies inversely with the square of the distance from the center of the balloon.
(b) Electric intensity E inside the balloon (r < R)
Inside the balloon, the electric field is zero. This is because the electric field is due to the charges on the surface of the balloon, and there are no charges inside the balloon.
(c) Electric potential V outside the balloon (r > R)
The electric potential V at a distance r from the center of the balloon is given by:
$$V=\frac{1}{4\pi\varepsilon_0}\int\frac{dq}{r}$$
Since the charge is uniformly distributed on the surface of the balloon, we can write dq = σ⋅dA, where σ is the surface charge density and dA is an element of area on the surface. The total charge on the balloon is q = σ⋅4πR², where R is the radius of the balloon. Substituting these into the equation for V, we get:
$$V=\frac{1}{4\pi\varepsilon_0}\int_S \frac{\sigma dA}{r}$$
$$V=\frac{1}{4\pi\varepsilon_0}\frac{\sigma}{r}⋅\int_S dA$$
$$V=\frac{1}{4\pi\varepsilon_0}\frac{\sigma}{r}⋅4πR²$$
$$V=\frac{\sigma R}{\varepsilon_0}\frac{1}{r}$$
This is the expression for the electric potential outside the balloon. It varies inversely with the distance from the center of the balloon.
(d) Electric potential V inside the balloon (r < R)
Inside the balloon, the electric potential is constant and is given by:
$$V=\frac{1}{4\pi\varepsilon_0}\int_0^R \frac{\sigma dA}{r}$$
$$V=\frac{1}{4\pi\varepsilon_0}\frac{\sigma}{r}⋅4πr²$$
$$V=\frac{\sigma R}{\varepsilon_0}$$
This is the expression for the electric potential inside the balloon. It is constant and does not depend on the distance from the center of the balloon.
