Let V be the electric potential at a given point. Then the electric field Ex along x -direction at that point is given by :

Question

Let V be the electric potential at a given point. Then the electric field Ex along x -direction at that point is given by : (A)  ∫ limits∞0 Vdx  (B)  (dV/dx)  (C)  -(dV/dx)  (D)  -V (dV/dx)

Tardigrade Admin · Accepted Answer

The electric field intensity at a point in an electric field in a given direction is equal to he negative potential gradient in that direction
i.e.,

E = - \frac{d V}{d x}

The negative sign signifies that the potential decreases in the direction of the electric field.

Q. Let $V$ be the electric potential at a given point. Then the electric field $E_{x}$ along $x$ -direction at that point is given by :

$0 \int \infty V d x$

$\frac{d V}{d x}$

$- \frac{d V}{d x}$

$- V \frac{d V}{d x}$

The electric field intensity at a point in an electric field in a given direction is equal to he negative potential gradient in that direction
i.e., $E = - \frac{d V}{d x}$
The negative sign signifies that the potential decreases in the direction of the electric field.

Q. Let V be the electric potential at a given point. Then the electric field Ex​ along x -direction at that point is given by :

0∫∞​Vdx

dxdV​

−dxdV​

−VdxdV​