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)  ∫0∞ Vdx  (B)  (dV/dx)  (C)  -(dV/dx)  (D)  -V(dV/dx)

Tardigrade Admin · Accepted Answer

The electric Held intensity at a point in an electric field in a given direction is equal to he negative potential gradient in that direction ie, E=-(dV/dx) The negative sign signifies that the potential decreases in the direction of the electric field. Note: It is evident from above relation, that the electric field E can also be expressed in the unit volt / metre or Vm-1 Thus, 1 NC-1=1 Vm-1

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

$\int_{0}^{\infty} V d x$

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

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

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

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​

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

$\int_{0}^{\infty} V d x$

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

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

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