A uniform electric field $$\mathrm{E}=(8 \mathrm{~m} / \mathrm{e}) \,\mathrm{V} / \mathrm{m}$$ is created between two parallel plates of length $$1 \mathrm{~m}$$ as shown in figure, (where $$\mathrm{m}=$$ mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of $$2 \mathrm{~m} / \mathrm{s}$$. The angle of the deviation $$(\theta)$$ of the path of the electron as it comes out of the field will be _________.

A charge of $$4 \,\mu \mathrm{C}$$ is to be divided into two. The distance between the two divided charges is constant. The magnitude of the divided charges so that the force between them is maximum, will be :

Two identical positive charges $$Q$$ each are fixed at a distance of '2a' apart from each other. Another point charge $$q_{0}$$ with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge $$\mathrm{q}_{0}$$ executes $$\mathrm{SHM}$$. The time period of oscillation of charge $$\mathrm{q}_{0}$$ will be :

An electron (mass $$\mathrm{m}$$) with an initial velocity $$\vec{v}=v_{0} i\left(v_{0}>0\right)$$ is moving in an electric field $$\vec{E}=-E_{0} \hat{i}\left(E_{0}>0\right)$$ where $$E_{0}$$ is constant. If at $$\mathrm{t}=0$$ de Broglie wavelength is $$\lambda_{0}=\frac{h}{m v_{0}}$$, then its de Broglie wavelength after time t is given by