## Related questions with answers

Two long parallel plates are separated by a distance of 15.0 cm. The bottom plate is kept at a potential of −250 V and the top at

$+ 250 \mathrm { V }$

. A charge of

$- 2.00 \mu \mathrm { C }$

is placed at a point 3.00 cm from the bottom plate. a. Find the electric potential energy of the charge. The charge is then moved vertically up to a point 3.00 cm from the top plate. b. What is the electrical potential energy of the charge now? c. How much work was done on the charge?

Solution

Verified$\textbf{.a)}$ The potential at a distance $(s)$ from the bottom plate can be written as follows

$V=V_{\text{bottom}}+E\times s$

Where $(V_{\text{bottom}})$ is the electric potential of the bottom plate and $(E)$ is the electric field between the two plates, and we know that the electric field between the two plates is $(\Delta V/\Delta r)$, where $(\Delta r=15$ cm). Thus, the electric potential at a distance $(s)$ from the bottom plate is

$V=-250 \mathrm{~ V}+\frac{250\mathrm{~ V} -(-250 \mathrm{~ V})}{0.15 \mathrm{~ m}} \times (0.03 \mathrm{~ m})$

$V=-150 \mathrm{~ V}$

now that we have the potential at the position of the charge, we can calculate its electric potential energy as follows

$E_{p}=qV=(- 2 \times 10^{-6} \mathrm{~ C})\times (-150 \mathrm{~ V})=3\times 10^{-4} \mathrm{~ J}$

$\textbf{.b)}$ The charge is now at a distance of (12 cm) from the bottom plate, so the potential at this point can be determined as follows

$V=-250 \mathrm{~ V}+\frac{250\mathrm{~ V} -(-250 \mathrm{~ V})}{0.15 \mathrm{~ m}} \times (0.12 \mathrm{~ m})$

$V=150 \mathrm{~ V}$

and the electric potential energy is therefore

$E_{p}=qV=(- 2 \times 10^{-6} \mathrm{~ C})\times (150 \mathrm{~ V})=-3\times 10^{-4} \mathrm{~ J}$

$\textbf{.c)}$ The work equals the change in the potential energy

$W=\Delta E_{p}=-3\times 10^{-4} \mathrm{~ J}-3\times 10^{-4} \mathrm{~ J}=-6\times 10^{-4} \mathrm{~ J}$

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