The electric potential V is given as a function of distance x (metre) by V=(5x2+10x-4)V. Value of electric field at x=1 m is:

Question

The electric potential V is given as a function of distance x (metre) by V=(5x2+10x-4)V. Value of electric field at x=1 m is: (A)  -23 V/m  (B)  11 V/m  (C)  6 V/m  (D)  -20 V/m

Tardigrade Admin · Accepted Answer

E=-(dV/dx) Electric field E is given by E=-(dV/dx) Given, V=5x2+10x-4 ∴ E=(dV/dx)=10x+10 At x=1,E=-(10 × 1 + 10) =-20 V/m

Q. The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5 x^{2} + 10 x - 4) V .$ Value of electric field at $x = 1 m$ is:

$- 23 V / m$

$11 V / m$

$6 V / m$

$- 20 V / m$

$E = - \frac{d V}{d x}$
Electric field $E$ is given by
$E = - \frac{d V}{d x}$
Given, $V = 5 x^{2} + 10 x - 4$
$∴ E = \frac{d V}{d x} = 10 x + 10$
At $x = 1, E = - (10 \times 1 + 10)$
$= - 20 V / m$

Q. The electric potential V is given as a function of distance x (metre) by V=(5x2+10x−4)V. Value of electric field at x=1m is:

−23V/m

11V/m

6V/m

−20V/m