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ELECTRIC CHANGES AND FIELDS

Physics
Inter Second Year
English
Short Answer Questions
Q1. Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole.

Answer

Ans. Let the point $P$ be at distance $r$ from the centre of the dipole $O$ as shown in figure.

RainbowThen $$\mathbf{E}_{_{+q}}=\mathbf{E}_{_{-q}}=\frac{1}{4\pi\varepsilon_{0}}\:\frac{q}{(r^{2}+a^{2})}$$ The total fleld at P is $$\mathbf{E}=-(\mathbf{E}_{+q}+\mathbf{E}_{-q})\cos\theta$$$$\mathbf{E}=\frac{-2q}{4\:\pi\:\varepsilon_{0}}\biggl[\:\frac{1}{(r^{2}+a^{2})}\biggr]\cos\theta$$where $\cos\theta=\biggl[\:\frac{a}{\sqrt{(r^{2}+a^{2})} }\biggr]$ $$\mathbf{E}=\frac{-2q}{4\:\pi\:\varepsilon_{0}}\biggl[\:\frac{1}{(r^{2}+a^{2})}\biggr]\biggl[\:\frac{a}{\sqrt{(r^{2}+a^{2})} }\biggr]$$$$\mathbf{E}=\frac{-2q}{4\:\pi\:\varepsilon_{0}}\biggl[\:\frac{a}{(r^{2}+a^{2})^{3/2}}\biggr] $$For $r>>a$ $$\mathbf{E}=\frac{-2\:q\:a}{4\pi\varepsilon_{0}r^{3}}$$$$\mathbf{E}=\frac{-P}{4\pi\varepsilon_{0} r^{3}}$$$$\text{Vector form}\:\: \mathbf{\vec{E}}=\frac{-P}{4\pi\varepsilon_{0} r^{3}}\hat{\mathbf{p}}$$