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

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Answer 1

$V (x) = 5 - 4 x^{2}$
$V (1) = 5 - 4 = 1 V, V (2) = 5 - 4 (2^{2}) = - 11 V$
Potential difference between $x = 1 m$ and $x = 2 m$ is $V_{12} = V_{1} - V_{2} = 1 - (- 11) = 12 V$
here, $E_{x} = - \frac{d V}{d x} = 8 x, E_{y} = - \frac{d V}{d y} = 0$ and $E_{z} = - \frac{d V}{d z} = 0$

The electric force on $1$ coulomb charge at $x = 1$ is $F = q E_{x} = 1 (8) = 8 N$

Answer 2

$V (x) = 5 - 4 x^{2}$

$V (1) = 5 - 4 = 1 V, V (2) = 5 - 4 (2^{2}) = - 11 V$

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

here, $E_{x} = - \frac{d V}{d x} = 8 x, E_{y} = - \frac{d V}{d y} = 0$ and $E_{z} = - \frac{d V}{d z} = 0$

The electric force on $1$ coulomb charge at $x = 1$ is $F = q E_{x} = 1 (8) = 8 N$