Question 788: Electrostats & Capacitance

Question

The electric potential ' V ' is given as a function of distance ' $x$ ' (metre) by $V=\left(4 x^2+8 x-3\right) V$. The value of electric field at $x=0.5 \mathrm{~m}$, in $\mathrm{V} / \mathrm{m}$ is

Accepted Answer

Given:
Electric potential,
$$ V = 4x^2 + 8x - 3 $$
The relation between electric field $E$ and potential $V$ (along $x$-axis) is:
$$ E = -\frac{dV}{dx} $$
Step 1: Differentiate $V$ with respect to $x$
$ \frac{dV}{dx} = \frac{d}{dx}(4x^2 + 8x - 3) $
Use rules of differentiation:
$\frac{d}{dx}(x^2) = 2x$
$\frac{d}{dx}(x) = 1$
$\frac{d}{dx}(	ext{constant}) = 0$
So,
$ \frac{dV}{dx} = 4 	imes 2x + 8 	imes 1 + 0 $
$ \frac{dV}{dx} = 8x + 8 $
Step 2: Write expression for $E$
$ E = -\frac{dV}{dx} = -(8x + 8) $
$ E = -8x - 8 $
Step 3: Substitute $x = 0.5$ m
$ E = -8(0.5) - 8 $
$ E = -4 - 8 $
$ E = -12 \, 	ext{V/m} $
Final Answer:
$ \boxed{-12~	ext{V/m}} $
Option
B
is correct.

Answer

-16

Answer

-12

Answer

0

Answer

+12

Electrostats & Capacitance

The electric potential ' V ' is given as a function of distance ' x ' (metre) by V=\left(4 x^2+8 x-3\right) V. The value of electric field at x=0.5 \mathrm{~m}, in \mathrm{V} / \mathrm{m} is

Choose the correct answer: