# A 2 V battery is connected to two parallel plates a distance 55 cm apart. If a -4.30 xx 10^-4 C charge was placed in the electric field, what force would it feel?

## What work is needed to move the $- 4.30 \times {10}^{-} 4 C$ charge from 0.138 m from the ground plate to 0.413 m from the ground plate?

Jun 21, 2017

Electric field is given by

$\vec{E} = \frac{V}{\mathrm{dh}} a t r$
direction of $\hat{r}$ is from positive plate to negative plate.

Inserting given values we get

$| \vec{E} | = \frac{2}{0.55} V {m}^{-} 1$ .....(1)

Force experienced by a charged particle $q$ in an electric field is

$\vec{F} = q \vec{E}$

Hence , using (1) we get

$\vec{F} = \frac{2}{0.55} \times \left(- 4.30 \times {1}^{-} 4\right) \hat{r}$
$| \vec{F} | = 1.56 \times {10}^{-} 3 N$
Direction is opposite to that of $\hat{r}$, i.e., towards positive plate

Work done in moving the charge is

$W = \vec{F} \cdot \vec{s}$

As the angle between displacement and force vectors is ${0}^{\circ}$,
$\cos \theta$ in the dot product $= \cos {0}^{\circ} = 1$

$\therefore W = | \vec{F} | | \vec{s} |$
$\implies W = 1.56 \times {10}^{-} 3 \times \left(0.413 - 0.138\right)$
$\implies W = 1.56 \times {10}^{-} 3 \times \left(0.413 - 0.138\right)$
$\implies W = 4.3 \times {10}^{-} 4 J$