(a) I am a little confused about this part. The point at A to B isn't radial. The electric field is radially outward, but if I look at the integral

$$\int_{a}^{b}\mathbf{E}\cdot d\mathbf{s} = \int_{a}^{b}\frac{\rho r}{3\epsilon_0}\mathbf{\hat{r}}\cdot d\mathbf{s}$$

The vector ds and r can't be in the same direction, so do I have to express it in norm form of the dot product? I am afraid to do so. So my wishful thinking answer (since it says it is 2 marks ) is

$$\int_{a}^{b}\mathbf{E}\cdot d\mathbf{s} = \int_{r}^{R} \frac{\rho r}{3\epsilon_0}dr$$

(b) Okay this one isn't too bad, but i am extremely paranoid. So I went back to the definition of potential

$$V = k\int\frac{dq}{d}$$ Since the density is uniform, I simply get $V = \dfrac{kQ}{d}$

Now I just substitute $r$ into equation and get $V = \dfrac{kQ}{r} = \dfrac{Q}{4\pi \epsilon_0r}$.

Note that "d" is the radial distance. I avoided using r or R becuase the picture uses r and R

Thank you for reading

## Best Answer

Why are you looking for a radial surface..? Look it as an Equipotential surface (a surface where all points are at same constant electric potential) as it comes with sphere. Hence, you can assume the points A to B as radial to find the potential difference.