Question

The electric potential $V$ (in volt) varies with $x$ (in metre) according to the relation $V=\left(5+4 x^{2}\right)$ The force experienced by a negative charge of $2 \times 10^{-6} C$ located at $x=0.5\, m$ is

• A. $2 \times 10^{-6} N$
• B. $4 \times 10^{-6} N$
• C. $6 \times 10^{-6} N$
• D. $8 \times 10^{-6} N$

Answer 1

Answer: (D)

Given, $V =5+4 x^{2}$
$\frac{d V}{d r}=E$
$\therefore \frac{d V}{d x}=8 x=E$
or $E=8 \times 0.5=4\, Vm ^{-1}$
Now, $F =q E$
$=2 \times 10^{-6} \times 4$
$=8 \times 10^{-6} N$