(B) The electric field is given as $\vec{E} = 20\hat{i}\, N/C$.For a uniform electric field,the potential difference between two points $A$ and $B$ is

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(B) The electric field is given as $\vec{E} = 20\hat{i}\, N/C$.For a uniform electric field,the potential difference between two points $A$ and $B$ is given by the relation $\Delta V = V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}$.Here,the displacement vector $\vec{r}_{AB} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} = (6 - 4)\hat{i} + (5 - 2)\hat{j} = 2\hat{i} + 3\hat{j}$.Substituting the values,we get $V_B - V_A = -(20\hat{i}) \cdot (2\hat{i} + 3\hat{j})$.$V_B - V_A = -20 \times 2 = -40\, V$.

Class 12 Physics Electric Potential and Capacitance Relation between Electric Field and Potential and Potential Gradient

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$A$ uniform electric field of $20\, N/C$ exists along the $x$-axis in a space. The potential difference $(V_B - V_A)$ for the points $A(4\,m, 2\,m)$ and $B(6\,m, 5\,m)$ is.....$V$.

A
$20 \sqrt{13}$
B
$-40$
C
$+40$
D
$-20 \sqrt{13}$

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Electric Potential and Capacitance

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Relation between Electric Field and Potential and Potential Gradient

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$A$ uniform electric field of $20\, N/C$ exists along the $x$-axis in a space. The potential difference $(V_B - V_A)$ for the points $A(4\,m, 2\,m)$ and $B(6\,m, 5\,m)$ is.....$V$.

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The potential of the electric field produced by a point charge at any point $(x, y, z)$ is given by $V = 3x^2 + 5$,where $x, y$ are in metres and $V$ is in volts. The intensity of the electric field at $(-2, 1, 0)$ is

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