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

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(A) The relationship between electric field $E$ and electric potential $V$ is given by the formula $E = -\frac{dV}{dx}$.Given $V = 5x^2 + 10x - 9$.Now,differentiate $V$ with respect to $x$:$\frac{dV}{dx} = \frac{d}{dx}(5x^2 + 10x - 9) = 10x + 10$.Substituting this into the electric field formula:$E = -(10x + 10) = -10x - 10$.To find the electric field at $x = 1 \text{ m}$,substitute $x = 1$ into the expression:$E = -10(1) - 10 = -10 - 10 = -20 \text{ V/m}$.

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

Easy

The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5x^2 + 10x - 9) \text{ V}$. The value of the electric field at $x = 1 \text{ m}$ is ...... $V/m$.

A
-$20$
B
$6$
C
$11$
D
-$23$

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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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MediumKCET 2019

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The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5x^2 + 10x - 9) \text{ V}$. The value of the electric field at $x = 1 \text{ m}$ is ...... $V/m$.

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The potential difference between two parallel plates is $10^4 \,V$. If the plates are separated by $0.5 \,cm$, the force on an electron between the plates is

$A$ uniform electric field of $400 \, V/m$ is directed $45^\circ$ above the $x$-axis. The potential difference $V_A - V_B$ is: (Coordinates are in cm) (in $, V$)

Two metal plates are separated by $2 \,cm$. The potentials of the plates are $-10 \,V$ and $+30 \,V$. The electric field between the two plates is: (in $\,V/m$)

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