Class 12PhysicsElectric Potential and CapacitanceRelation between Electric Field and Potential and Potential Gradient
EasyWrite the relation between the electric field of an electric charge and electrostatic potential at any point.
(N/A) The electric field $\vec{E}$ at any point is related to the electrostatic potential $V$ at that point by the negative gradient of the potential.
Mathematically,this is expressed as: $\vec{E} = -\nabla V$.
In Cartesian coordinates,this can be written as: $\vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
This relation signifies that the electric field points in the direction of the steepest decrease of the electrostatic potential.
Mathematically,this is expressed as: $\vec{E} = -\nabla V$.
In Cartesian coordinates,this can be written as: $\vec{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
This relation signifies that the electric field points in the direction of the steepest decrease of the electrostatic potential.
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If potential (in volts) in a region is expressed as $V(x, y, z) = 6xy - y + 2yz$,the electric field (in $N/C$) at point $(1, 1, 0)$ is:
The figure shows a set of equipotential surfaces. The magnitude and direction of the electric field that exists in the region is .........
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View SolutionThe electric potential $V(x, y, z)$ for a planar charge distribution is given by:
$V(x, y, z) = \begin{cases} 0 & \text{for } x < -d \\ -V_0(1 + \frac{x}{d})^2 & \text{for } -d \le x < 0 \\ -V_0(1 + 2\frac{x}{d}) & \text{for } 0 \le x < d \\ -3V_0 & \text{for } x \ge d \end{cases}$
where $-V_0$ is the potential at the origin and $d$ is a distance. The graph of the electric field as a function of position is:
$V(x, y, z) = \begin{cases} 0 & \text{for } x < -d \\ -V_0(1 + \frac{x}{d})^2 & \text{for } -d \le x < 0 \\ -V_0(1 + 2\frac{x}{d}) & \text{for } 0 \le x < d \\ -3V_0 & \text{for } x \ge d \end{cases}$
where $-V_0$ is the potential at the origin and $d$ is a distance. The graph of the electric field as a function of position is:
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View SolutionIf $V = -5x + 3y + \sqrt{15}z$,then find the magnitude of the electric field $E(x, y, z)$ in units.
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View SolutionThe electric field in a region is given by $\vec{E} = (Ax + B)\hat{i}$ where $E$ is in $N\,C^{-1}$ and $x$ is in meters. The values of constants are $A = 20\, SI\, \text{unit}$ and $B = 10\, SI\, \text{unit}$. If the potential at $x = 1$ is $V_1$ and that at $x = -5$ is $V_2$, then $V_1 - V_2$ is.....$V$.
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