Question

A graph of the $$x$$ component of the electric field as a function of $$x$$ in a region of space is shown in the above figure. The scale of the vertical axis is set by $$E_{xs} = 20.0 \ N/C$$. The $$y$$ and $$z$$ components of the electric field are zero in this region. If the electric potential at the origin is $$10 \ V$$,
(a) What is the electric potential at $$x=2.0 \ m$$,
(b) What is the greatest positive value of the electric potential for points on the $$x$$ axis for which $$0 \leq x \leq 6.0 \ m$$, and
(c) for what value of $$x$$ is the electric potential zero?

Answer 1

(a) By Eq. $$ \ V_f-V_i=- \int^f_i \overset{\rightarrow}{E}.\overset{\rightarrow}{ds}$$, the change in potential is the negative of the "area" under the curve. Thus, using the area-of-a-triangle formula, we have

$$ V-10=- \int^{x=2}_0 \overset{\rightarrow}{E}.\overset{\rightarrow}{ds} = \dfrac{1}{2} (2)(20)$$

which yields $$V=30 \ V$$

(b) For any region within $$0<x<3m, - \int \overset{\rightarrow}{E}.\overset{\rightarrow}{ds}$$ is positive, but for any region for which $$x>3 \ m$$ it is negative. Therefore, $$V=V_{max}$$ occurs at $$x=3 \ m$$

$$ V-10=- \int^{x=3}_0 \overset{\rightarrow}{E}.\overset{\rightarrow}{ds} = \dfrac{1}{2} (3)(20) $$

which yields $$V_{max} = 40 \ V$$.

(c) In view of our result in part (b), we see that now (to find $$V=0$$) we are looking for some $$X>3 \ m$$ such that the "area" from $$x=3 \ m$$ to $$x=X$$ is $$40 \ V$$. Using the formula for a triangle $$(3<x<4)$$ and a rectangle $$(4<x<X)$$, we require

$$ \dfrac{1}{2}(1)(20) + (X-4)(20) = 40 $$

Therefore, $$X=5.5 \ m$$.