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Question

(B) The electric field $E$ between two large parallel conducting plates is uniform and is given by $E = \frac{V}{d}$,where $V$ is the potential differ

Vedclass · Accepted Answer

$\frac{2}{5} V$

Vedclass · Answer

(B) The electric field $E$ between two large parallel conducting plates is uniform and is given by $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the separation between the plates.Given $d = 10 \ cm$,so $E = \frac{V}{10 \ cm}$.The potential difference between two points is given by $\Delta V = E \cdot \Delta x$,where $\Delta x$ is the projection of the displacement vector between the points along the direction of the electric field.The electric field is directed from the positive plate to the negative plate (horizontally).The horizontal distance (projection) between points $A$ and $B$ is the same as the horizontal distance between $C$ and $B$,which is $4 \ cm$.Therefore,the potential difference between $A$ and $B$ is $V_{AB} = E \cdot (4 \ cm) = \left( \frac{V}{10 \ cm} ight) 	imes 4 \ cm = \frac{4}{10} V = \frac{2}{5} V$.

Two large plane parallel conducting plates are kept $10 \ cm$ apart as shown in the figure. The potential difference between them is $V$. The potential difference between the points $A$ and $B$ (shown in the figure) is

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