# Electrostatic potential energy

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# The electric potential varies in space according to the relation $V=3x+4y.$ A particle of mass 0.1 kg starts from rest from point (2,3.2) under the influence of this field. The charge on the particle is $+1\text{\hspace{0.17em}\hspace{0.17em}}\mu C.$ Assume V and (x, y) are in SI units.

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## ${E}_{x}=-\frac{\delta V}{\delta x}=-3\text{\hspace{0.17em}\hspace{0.17em}}V{m}^{-1};\text{\hspace{0.17em}\hspace{0.17em}}{E}_{y}=-\frac{\delta V}{\delta y}=-4\text{\hspace{0.17em}\hspace{0.17em}}V{m}^{-1}$${a}_{x}=\frac{q{E}_{x}}{m}=-\frac{1×{10}^{-6}×3}{0.1}=-3×{10}^{-5}\text{\hspace{0.17em}\hspace{0.17em}}m{s}^{-2}$${a}_{y}=\frac{q{E}_{y}}{m}=-\frac{1×{10}^{-6}×4}{0.1}=-4×{10}^{-5}\text{\hspace{0.17em}\hspace{0.17em}}m{s}^{-2}$Time taken to cross the x-axisUsing $s=ut+\frac{1}{2}a{t}^{2}\text{\hspace{0.17em}\hspace{0.17em}}we\text{\hspace{0.17em}\hspace{0.17em}}get$$3.2=\frac{1}{2}×4×{10}^{-5}\text{\hspace{0.17em}\hspace{0.17em}}{t}^{2}$or $t=400\text{\hspace{0.17em}\hspace{0.17em}}s$${v}_{x}={a}_{x}t=-3×{10}^{-5}×400=12×{10}^{-3}\text{\hspace{0.17em}\hspace{0.17em}}m{s}^{-1}$${v}_{y}={a}_{y}t=-4×{10}^{-5}×400=16×{10}^{-3}\text{\hspace{0.17em}\hspace{0.17em}}m{s}^{-1}$$v=\sqrt{{v}_{x}^{2}+{v}_{y}^{2}}=20×{10}^{-3}\text{\hspace{0.17em}\hspace{0.17em}}m{s}^{-1}$

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Charge Q is given a displacement $\stackrel{\to }{\mathrm{r}}=\mathrm{a}\stackrel{^}{\mathrm{i}}+\mathrm{b}\stackrel{^}{\mathrm{j}}$ in an electric field $\stackrel{\to }{\mathrm{E}}={\mathrm{E}}_{1}\stackrel{^}{\mathrm{i}}+{\mathrm{E}}_{2}\stackrel{^}{\mathrm{j}}$. The work done is