If on the x-axis electric potential decreases | vedclass

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(B) The electric field component along the $x$-axis is given by $E_x = -\frac{dV}{dx}$.Given that the potential decreases uniformly from $60 \,V$ to $

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May be greater than $10 \,V/m$

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(B) The electric field component along the $x$-axis is given by $E_x = -\frac{dV}{dx}$.Given that the potential decreases uniformly from $60 \,V$ to $20 \,V$ over a distance of $4 \,m$ (from $x = -2 \,m$ to $x = +2 \,m$),the magnitude of the $x$-component of the electric field is $|E_x| = \frac{\Delta V}{\Delta x} = \frac{60 \,V - 20 \,V}{2 \,m - (-2 \,m)} = \frac{40 \,V}{4 \,m} = 10 \,V/m$.Since the potential is only given as a function of $x$,we cannot conclude that the electric field components in the $y$ or $z$ directions are zero. The total electric field is $E = \sqrt{E_x^2 + E_y^2 + E_z^2}$.Even if $E_y$ and $E_z$ are non-zero,the magnitude of the electric field $E$ will be $\sqrt{(10)^2 + E_y^2 + E_z^2}$,which is greater than or equal to $10 \,V/m$.Therefore,the magnitude of the electric field at the origin may be greater than $10 \,V/m$.

If on the $x$-axis electric potential decreases uniformly from $60 \,V$ to $20 \,V$ between $x=-2 \,m$ to $x=+2 \,m$,then the magnitude of electric field at the origin

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