Physics - Electrostatics Question with Solution | TestHub

According to conservation of energy, we get $K_a+U_a=K_b+U_b$ Here, $K_a=0$ and the potential energies are $U_a=q^{\prime }{V}_a$ and $U_b=q^{\prime }{V}_b$ $$\begin{array}{rl}& \therefore 0+q^{\prime }{V}_a=\frac{1}{2}m{v}^2+q^{\prime }{V}_b\\ & \text{ or }v=\sqrt{\frac{2q^{\prime }\left(V_a-V_b\right)}{m}}\\ & V_a=\left(9.0\times 10^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2}\right)\end{array}$$ $$\begin{array}{r}\left(\frac{3\times 10^{-9}\mathrm{C}}{0.01\mathrm{m}}+\frac{-3\times 10^{-9}\mathrm{C}}{0.02\mathrm{m}}\right)=1350\mathrm{V}\\ V_b=\left(9.0\times 10^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2}\right)\\ \left(\frac{3\times 10^{-9}\mathrm{C}}{0.02\mathrm{m}}+\frac{-3\times 10^{-9}\mathrm{C}}{0.01\mathrm{m}}\right)=-1350\mathrm{V}.\end{array}$$ $$\begin{array}{rl}\therefore & v=\sqrt{\frac{2\left(2\times 10^{-9}\mathrm{C}\right)(2700\mathrm{V})}{5\times 10^{-3}\mathrm{kg}}}=4.65\times 10^{-2}\mathrm{m}{\mathrm{s}}^{-1}\\ & =4.65\mathrm{cm}{\mathrm{s}}^{-1}\end{array}$$