(D) The relationship between electric field $(E)$ and electric potential $(V)$ is given by the formula $E = -\frac{dV}{dx}$.Given $V = 5x^2 + 10x - 9$

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(D) The relationship between electric field $(E)$ and electric potential $(V)$ is given by the formula $E = -\frac{dV}{dx}$.Given $V = 5x^2 + 10x - 9$.Differentiating $V$ with respect to $x$:$\frac{dV}{dx} = \frac{d}{dx}(5x^2 + 10x - 9) = 10x + 10$.Therefore,$E = -(10x + 10)$.At $x = 1 \, m$:$E = -(10(1) + 10) = -(10 + 10) = -20 \, V/m$.

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

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The electric potential $(V)$ as a function of distance $(x)$ [in meters] is given by $V = (5x^2 + 10x - 9) \, V$. The value of the electric field at $x = 1 \, m$ would be ...... $V/m$.

A
$20$
B
$6$
C
$11$
D
$-20$

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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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The electric field at a distance $x$ from the origin is given by $E = 100/x^2$. The potential difference between the points at $x = 10 \, m$ and $x = 20 \, m$ is ...... $V$.

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The electric potential $(V)$ as a function of distance $(x)$ [in meters] is given by $V = (5x^2 + 10x - 9) \, V$. The value of the electric field at $x = 1 \, m$ would be ...... $V/m$.

Similar Questions

Due to some charges on the $x$-axis,the potential at a point on the $x$-axis is given by $V(x) = 20/(x^2 - 4) \ V$. The electric field at $x = 4 \ \mu m$ is given by:

The electrostatic potential inside a charged spherical ball is given by $V = b - ar^2$,where $r$ is the distance from the centre; $a$ and $b$ are constants. Then,the charge density inside the ball is:

The variation of potential with distance $R$ from a fixed point is as shown in the graph. The electric field at $R = 5\,m$ is......$V/m$.

In a certain region,the electric potential is given by $V = 6x - 8xy^2 - 8y + 6yz - 4z^2 \, V$. The magnitude of the force on a charge of $2 \, C$ at the origin is ........ $N$.

The electric field at a distance $x$ from the origin is given by $E = 100/x^2$. The potential difference between the points at $x = 10 \, m$ and $x = 20 \, m$ is ...... $V$.

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