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1. Electric charges and fields — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics1. Electric charges and fields1 Mark
MCQ
The electrostatic force of interaction between an uniformly charged rod having total charge $Q$ and length $L$ and a point charge $q$ as shown in figure is
  • ✓
    $\frac{1}{{4\pi { \in _0}}}\frac{{qQ}}{{d(d + L)}}$
  • B
    $\frac{1}{{4\pi { \in _0}}}\frac{4{qQ}}{{(2d + L)^2}}$
  • C
    $\frac{1}{{4\pi { \in _0}}}\frac{{Qq}}{{d^2}}$
  • D
    $\frac{1}{{4\pi { \in _0}}}\frac{{qQ}}{{{{(d + L)}^2}}}$
✓

Answer

Correct option: A.
$\frac{1}{{4\pi { \in _0}}}\frac{{qQ}}{{d(d + L)}}$
a
$\int {{\rm{dF}}}  = \int_0^{{\rm{d}} - {\rm{L}}} {\frac{1}{{4\pi { \in _0}}}} \frac{{\left( {\frac{{\rm{Q}}}{{\rm{L}}}{\rm{dx}}} \right){\rm{q}}}}{{{{\rm{x}}^2}}}$

$\Rightarrow \mathrm{F}=\frac{1}{4 \mathrm{n} \in_{0}} \frac{\mathrm{qQ}}{\mathrm{L}}\left|\frac{1}{-\mathrm{x}}\right|_{\mathrm{d}}^{\mathrm{d}+\mathrm{L}}$

$F = \frac{1}{{4\pi { \in _0}}}\frac{{Qq}}{L}\left[ { - \frac{1}{{d + L}} + \frac{1}{d}} \right]$

$ = \frac{1}{{4\pi { \in _0}}}\frac{{qQ}}{L}\left[ {\frac{{ - d + d + L}}{{d(d + L)}}} \right]$

${\rm{F}} = \frac{1}{{4\pi { \in _0}}}\frac{{{\rm{qQ}}}}{{{\rm{d}}({\rm{d}} + {\rm{L}})}}$

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