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Question

(B) In a uniform electric field $\vec{E}$,the potential $V$ at a position $\vec{r}$ relative to point $P$ is given by $V = V_P - \vec{E} \cdot \vec{r}

Vedclass · Accepted Answer

$589.2$

Vedclass · Answer

(B) In a uniform electric field $\vec{E}$,the potential $V$ at a position $\vec{r}$ relative to point $P$ is given by $V = V_P - \vec{E} \cdot \vec{r} = V_P - Er \cos 	heta$,where $	heta$ is the angle between the radius vector and the field direction.The potential varies between $V_{min} = V_P - Er$ (at $	heta = 0^o$) and $V_{max} = V_P + Er$ (at $	heta = 180^o$).Given $V_{min} = 589.0\,V$ and $V_{max} = 589.8\,V$,the potential difference across the diameter is $2Er = 589.8 - 589.0 = 0.8\,V$. Thus,$Er = 0.4\,V$.The center potential $V_P$ is the average: $V_P = (589.0 + 589.8) / 2 = 589.4\,V$.For a point at an angle $	heta = 60^o$,the potential is $V = V_P - Er \cos(60^o) = 589.4 - 0.4 	imes 0.5 = 589.4 - 0.2 = 589.2\,V$.

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