(a) Figure 24-10a shows two points i and fin a uniform electric field `overset(to) (E )`. The points lie on the same electric field line (not shown) and are separated by a distanced. Find the potential difference `V_(f)- V_(i)` by moving a positive test charge `q_(0)` from i to f along the path shown, which is parallel to the field direction.
(b) Now find the potential difference `V_(f)- V_(i)` by moving the positive test charge `q_0` from `i` to `f` along the path `icf` shown in Fig. 24-10b.

Explanation

We can find the potential difference between any two points in an electric field by integrating `overset(to) ( E) .d overset(to) (s)` along a path connecting those two points according to Eq. 24-20.
Calculations: We have actually already done the calculation for such a path in the direction of an electric field line in a uniform field when we derived Eq. 24-23. With slight changes in notation, Eq. 24-23 gives us
`V_(f)- V_(i)= - Ed.` (Answer)
Calculations: The Key Idea of (a) applies here too, except now we move the test charge along a path that consists of two lines: is and cf. .At all points along line is, the displacement `d overset(to)s` of the test charge is perpendicular to `overset(to) (E)`. Thus, the angle `theta` between `overset(to)(E) and d overset(to)(s)` is 90°, and the dot product `overset(to)(E). d overset(to)(s)` is 0. Equation 24-20 then tells us that points i and c are at the same potential: `V_(c ) - V_(i)= 0`. Ah, we should have seen this coming. The points are on the same equipotential surface, which is perpendicular to the electric field lines.
For lines `cf` we have `theta=45^@` and, from Eq `24-20`,
`V_(f)- V_(i)-int_(c)^(f) overset(to)( E).doverset(to)(s)=-int_(c)^(f) E ( cos 45^@) ds`
`=- E ( cos 45^@) int_(c )^(f) ds`.