Q147 / 160 · AP EAMCET 2021 (23 Aug Shift 1) · Physics

Two opposite charges each of magnitude $500 \mu \mathrm{C}$ are $10 \mathrm{~cm}$ apart. Find electric field intensity at a distance of $25 \mathrm{~cm}$ from the midpoint on axial line of the dipole.

  • Correct answer
    A
    $5.76 \times 10^7 \mathrm{NC}^{-1}$
  • B
    $9.28 \times 10^7 \mathrm{NC}^{-1}$
  • C
    $13.1 \times 10^{10} \mathrm{NC}^{-1}$
  • D
    $20.5 \times 10^7 \mathrm{NC}^{-1}$

Solution

Given, charge of dipole, $q=500 \times 10^{-6} \mathrm{C}$ Separation between charges, $2 a=10 \mathrm{~cm}$ $$ =10 \times 10^{-2} \mathrm{~m} $$ Distance of location from mid point of axis, $$ r=25 \mathrm{~cm}=25 \times 10^{-2} \mathrm{~m} $$ $\because$ Electric field on axis of dipole, $$ E=\frac{4 k q a r}{\left(r^2-a^2\right)^2} $$ where, $k$ is Coulomb's constant $=9 \times 10^9 \mathrm{C}^2 \mathrm{~m}^{-2} \mathrm{~N}^{-1}$ $$ \begin{aligned} & 4 \times 9 \times 10^9 \times 500 \times 10^{-6} \\ & \Rightarrow \quad E=\frac{\times 5 \times 10^{-2} \times 25 \times 10^{-2}}{\left[\left(25 \times 10^{-2}\right)^2-\left(5 \times 10^{-2}\right)^2\right]^2} \\ & =\frac{2.25 \times 10^5}{[0.0625-0.0025]^2} \\ & =\frac{2.25 \times 10^5}{(0.06)^2} \\ & =\frac{2.25 \times 10^5}{3.6 \times 10^{-3}} \\ & =6.25 \times 10^7 \\ & \approx 6 \times 10^7 \mathrm{NC}^{-1} \\ & \end{aligned} $$ The result is close to option (a).

Hence, the correct answer is option: 'A'.

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