(D) The relationship between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.Given the potential function $V = 8x^2 + 2$.Differe

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(D) The relationship between electric field $E$ and potential $V$ is given by $E = -\frac{dV}{dx}$.Given the potential function $V = 8x^2 + 2$.Differentiating with respect to $x$,we get $\frac{dV}{dx} = \frac{d}{dx}(8x^2 + 2) = 16x$.Thus,the electric field $E_x = -16x$.At the point $(-4, 0)$,the $x$-coordinate is $x = -4$.Substituting this value into the expression for $E_x$,we get $E_x = -16(-4) = 64 \, V/m$.The magnitude of the electric field is $|E_x| = 64 \, V/m$.

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

Easy

In a certain region of space,the variation of potential with distance from the origin as we move along the $x$-axis is given by $V = 8x^2 + 2$,where $x$ is the $x$-coordinate of a point in space. The magnitude of the electric field at a point $(-4, 0)$ is .......... $V/m$.

A
$16$
B
$32$
C
$48$
D
$64$

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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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In a region,the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$,where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2 \ C$ situated at point $(1, 1, 1)$ is

MediumAIPMT 2014

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Write the relation between electric field and electrostatic potential.

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In a certain region of space,the variation of potential with distance from the origin as we move along the $x$-axis is given by $V = 8x^2 + 2$,where $x$ is the $x$-coordinate of a point in space. The magnitude of the electric field at a point $(-4, 0)$ is .......... $V/m$.

Similar Questions

$A$ uniform electric field having a magnitude $E_0$ and direction along the positive $X$-axis exists. If the potential $V$ is zero at $x = 0$,then its value at $x = +x$ will be:

If the potential energy of an electron and a proton at a distance $r$ is given by the formula $U = - \left( \frac{ke^2}{3r^3} \right)$,then which law of force applies?

Two plates are separated by a distance of $20 \, cm$. The potential difference between them is $10 \, V$. The electric field between the two plates is ...... $V m^{-1}$.

In a region,the potential is represented by $V(x, y, z) = 6x - 8xy - 8y + 6yz$,where $V$ is in volts and $x, y, z$ are in metres. The electric force experienced by a charge of $2 \ C$ situated at point $(1, 1, 1)$ is

Write the relation between electric field and electrostatic potential.

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