Question

The electric potential $V$ (n volts) varies with $x$ (in meter) according to the relation $V=5+4x^2$
The force experienced by a negative charge of $\mu C$ located at $x=0.5m$ 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: (A), (D)

The correct options are
A $2\times {10}^{-6}N$
D $8\times {10}^{-6}N$

$V=5+4x^2$

Now, we know that

$E_x=\frac{\partial v}{\partial x}=8x,Ey=\frac{\partial v}{\partial y}=0,E_z=\frac{\partial v}{\partial z}=0$

$\Rightarrow E=E_x=8x$

Now, force experienced by a charged

particle in an electric field is given by,

$F=qE$

$\Rightarrow F=-(2\times {10}^{-6})\times 8x$

At $x=0.5m$,

$F=(2\times {10}^{-6})\times (8\times 0.5)$

$\Rightarrow F=8\times {10}^{-6}N$