The electric field vec{E} between two points | vedclass

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(B) The potential difference between two points in a uniform electric field is given by $\Delta V = -\int \vec{E} \cdot d\vec{l}$.For a uniform electr

Vedclass · Accepted Answer

$Ed \cos 60^o$

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(B) The potential difference between two points in a uniform electric field is given by $\Delta V = -\int \vec{E} \cdot d\vec{l}$.For a uniform electric field,this simplifies to $\Delta V = E \cdot \Delta r$,where $\Delta r$ is the displacement along the direction of the electric field.In the given figure,the displacement from point $1$ to point $2$ is $d$ at an angle of $60^o$ with the electric field lines.The component of displacement $d$ along the direction of the electric field is $d \cos 60^o$.Therefore,the potential difference between points $1$ and $2$ is $V = E(d \cos 60^o) = Ed \cos 60^o$.

The electric field $\vec{E}$ between two points is constant in both magnitude and direction. Consider a path of length $d$ at an angle $\theta = 60^o$ with respect to the field lines as shown in the figure. The potential difference between points $1$ and $2$ is

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