Here, p = 1 0 − 5 C × 5 × 1 0 − 3 m = 5 × 1 0 − 8 C m E = 2. A sample of HCl gas is placed in a uniform electric field of magnitude 3 × 104 N C-1. • Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail, Electric Dipole : Solved Example Problems, The position vector for the +q on the positive x-axis is ai and position vector for the +q charge the negative x axis is -a, In this case one charge is placed at the origin, so its position vector is zero. Question: An electric dipole consists of a positive and negative charge of 4µC each placed at a distance of 5mm.Calculate dipole moment. From both cases (a) and (b), we can infer that in general the electric dipole moment depends on the choice of the origin and charge configuration. Calculate the electric dipole moment for the following charge configurations. It is characterised by dipole moment vector . It is because of this reason that the electric dipole moment of an electric dipole (total charge is zero) is always directed from –q to +q, independent of the choice of the origin. Electric Dipole moment: Electric dipole moment is defined as the product of magnitude of one of the charges and length of the dipole say ‘2a’ i.e. Note that in this case p is directed from -2q to +q. View 5 Recommendations \frac{a}{\sqrt{r^2 + a^2}} $, $\large E = 2 \frac{1}{4\pi \epsilon_0}\frac{q \times 2a}{(r^2 + a^2)^{3/2}} $, $\large E = 2 \frac{1}{4\pi \epsilon_0}\frac{p}{(r^2 + a^2)^{3/2}} $, $\large E = 2 \frac{1}{4\pi \epsilon_0}\frac{p}{r^3} $, Note : The angle between electric field Intensity & dipole moment is 180°, Electric field due to the dipole at point P (r,θ)is given by, $ \displaystyle \vec{E} = \frac{2p cos\theta}{4 \pi \epsilon_0 r^3}\, \hat{r} + \frac{p sin\theta}{4 \pi \epsilon_0 r^3} \, \hat{\theta} $ ; where r >> a. We serve exceptional features to students that differentiate us from the rest of the institutes.We provide physics and other subjects in Hindi also. Consider AB is an electric dipole having mid point O & length 2a . English Quiz For IBPS PO , SBI PO, IBPS CLERK, SBI CLERK , RRB , SSC, India Railways, DMRC, Quant Quiz For IBPS CLERK , SBI PO, IBPS PO, SBI CLERK, SSC, RRB, DMRC, IR, Reasoning Quiz for IBPS Clerk , SBI PO, IBPS PO, SBI Clerk, RRB, General (Tips, facts, education, science related), GS Quiz For IBPS CLERK , SBI PO, IBPS PO, SBI CLERK, SSC, RRB, DMRC, IR, Indian Festivals and Important date-Short essay, National Cancer Awareness Day – 7 November, CA Final Exam Eligibility , Exams , Subjects & Syllabus. Electric Dipole moment: Electric dipole moment is defined as the product of magnitude of one of the charges and length of the dipole say ‘2a’ i.e. Dipole moment is a vector quantity whose magnitude is equal to the product of the magnitude of charges forming the dipole and their separation. Electric dipole moment is measured in Coulomb-meters (C m) in the SI system. You may need to download version 2.0 now from the Chrome Web Store. Another way to prevent getting this page in the future is to use Privacy Pass. dipole moment is about 3.336 × 10 −12 C m, and a debye (D) is 10 −18 cgs esu. It is directed from – q to + q i.e.from negative charge to the positive charge. See the answer. SI unit of electric dipole moment isCoulomb- meter (Cm). Statement of a problem № 36591 next . Solution: The calculated dipole moment for this condition is, p = q x d. Thus, p = 2 x 0.02 = 0.04 C-m. Case (c) . Cloudflare Ray ID: 5ec48d1799072778 The dipole moment of each HCl molecule is 3.4 × 10-30 Cm. The direction of dipole moment is from negative to positive charge. The "internal" electric field vector, meanwhile, is parallel but in the opposite direction of both the dipole moment vector and the "external" electric field dipole. Dipole moment is a vector quantity whose magnitude is equal to the product of the magnitude of charges forming the dipole and their separation. • Electric force and Torque on an Electric dipole in a uniform electric field: For the practice of Electrostatics problems & more, click on CBSE NCERT Physics Class 12. We are in NEWS Case (b) In this case one charge is placed at the origin, so its position vector is zero. At a Point on axial Line : We have to find Electric field intensity at Point P . Solution. Electric field Intensity due to an electric dipole.
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