SATHEE: electric-charges-and-fields Question 16

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Question: Q. 6. Find the expression for electric field intensity in an axial position due to electric dipole.

R [O.D. Comptt. I, II, III, 2013]

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Solution:

Ans. Consider an electric dipole whose length is $2 a$ and centre at $O$ . From the mid-point $O$ , consider a point $P$ at a distance $r$ , where the electric field intensity is to be determined.

$\begin{aligned} We have & E = \frac{1}{4 π ε_{0}} \cdot \frac{q}{r^{2}} & E_{1} = \frac{1}{4 π ε_{0}} \cdot \frac{q}{(r - a)^{2}} & E_{2} = \frac{1}{4 π ε_{0}} \cdot \frac{q}{(r + a)^{2}} & E = E_{1} - E_{2} & E = \frac{q}{4 π ε_{0}} [\frac{1}{(r - a)^{2}} - \frac{1}{(r + a)^{2}}] \end{aligned}$

$\begin{aligned} E = \frac{q}{4 π ε_{0}} [\frac{(r + a)^{2} - (r - a)^{2}}{(r - a)^{2} \cdot (r + a)^{2}}] & E = \frac{q}{4 π ε_{0}} \cdot \frac{4 a r}{{(r^{2} - a^{2})}^{2}} \end{aligned}$

for, $r^{2} » > a^{2}$

$∴ \begin{aligned} E & = \frac{q}{4 π ε_{0}} | \frac{4 a r}{r^{4}} | E & = \frac{1}{4 π ε_{0}} \cdot \frac{2 \times q \times 2 a \times r}{r^{4}} E & = \frac{1}{4 π ε_{0}} \cdot \frac{2 \vec{p}}{r^{3}} N / C [∵ p = 2 q a] 2 \end{aligned}$

Commonly Made Error

Several candidates derivelan expression for intensity of electric field $E$ at a point in the broadside position i.e., coaxial position, instead of that in the endon position as required. Some candidates do not understand which derivation to be given. Hence they write both the derivations. Many candidates are notable to draw the correctly labelled diagram.

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Commonly Made Error

विषयसूची

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

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