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CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 13 of 27
Marks: +1, -0
SECTION -C
(a) Draw the equipotential surfaces corresponding to a uniform electric field in the ZZ-direction.
(b) Derive an expression for the electric potential at any point along the axial line of an electric dipole.
Solution:  
(a)Drawing of equipotential surfaces
(b) Derivation of the expression of electric potential
(b)
Potential at point PP
  Vp=V−q+V+q\; V p = V - q + V + q
  =  14πε0  −q(r+a)+  14πε0  q(r−a)\; = \; \frac{1}{4 \pi \varepsilon_0} \; \frac{-q}{(r+a)} + \; \frac{1}{4 \pi \varepsilon_0} \; \frac{q}{(r-a)}
  =  q4πε0[  1(r−a)−  1(r+a)]\; = \; \frac{q}{4 \pi \varepsilon_0} \left[ \; \frac{1}{(r-a)} - \; \frac{1}{(r+a)} \right]
  =  q4πε0[  r+a−r+a(r−a)(r+a)]\; = \; \frac{q}{4 \pi \varepsilon_0} \left[ \; \frac{r+a-r+a}{(r-a)(r+a)} \right]
  =  q4πε0×  2a(r2−a2)=  q×2a4πε0(r2−a2)\; = \; \frac{q}{4 \pi \varepsilon_0} \times \; \frac{2a}{(r^2 - a^2)} = \; \frac{q \times 2a}{4 \pi \varepsilon_0 (r^2 - a^2)}
  =  14πε0  p(r2−a2)\; = \; \frac{1}{4 \pi \varepsilon_0} \; \frac{p}{(r^2 - a^2)}
where, pp is the dipole moment.
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