Question

The electric potential in a certain region along the x-axis varies with x according to the relation $V\left(x\right)=5-4x^2$. Then, the correct statement is :

Answer 1

$V\left(x\right)=5-4x^2$

$V\left(1\right)=5-4=1V,V\left(2\right)=5-4\left(2^2\right)=-11V$

Potential difference between $x=1m$ and $x=2m$ is $V_{12}=V_1-V_2=1-(-11)=12V$

here, $E_x=-\frac{dV}{dx}=8x,E_y=-\frac{dV}{dy}=0$ and $E_z=-\frac{dV}{dz}=0$

The electric force on $1$ coulomb charge at $x=1$ is $F=q{E}_x=1\left(8\right)=8N$