The electric potential V is given as a function of distance x (m) by V=(5.2+10x-9) volt. Value of electric field at x = 1 m is

Question

The electric potential V is given as a function of distance x (m) by V=(5.2+10x-9) volt. Value of electric field at x = 1 m is (A) 20V/m (B) 6V/m (C) 11 V/m (D) -23V/m

Tardigrade Admin · Accepted Answer

Voltage V=5x2+10x-9 Electric field E=-(dV/dx)=-(d/dx)(5x2+10x-9) =-(10x-10) At x=1 text m ∴ E=10(1)+10=20 V/m

Q. The electric potential V is given as a function of distance $x$ (m) by $V = (5.^{2} + 10 x - 9)$ volt. Value of electric field at $x$ = 1 m is

20V/m

6V/m

11 V/m

-23V/m

Voltage $V = 5 x^{2} + 10 x - 9$ Electric field $E = - \frac{d V}{d x} = - \frac{d}{d x} (5 x^{2} + 10 x - 9)$ $= - (10 x - 10)$ At $x = 1 m$ $∴$ $E = 10 (1) + 10 = 20 V / m$

Q. The electric potential V is given as a function of distance x (m) by V=(5.2+10x−9) volt. Value of electric field at x = 1 m is