vector form of coulomb's law derivation

Hence Q should be divided in two equal parts. Fair =${1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}}$ and Fmedium = ${1 \over {4\pi {\varepsilon _0}{\varepsilon _r}}}{{{q_1}{q_2}} \over {{r^2}}}$, ${{{F_{medium}}} \over {{F_{air}}}} = {1 \over {{\varepsilon _r}}}$= K. εr or K = Dielectric constant or Relative permittivity or specific inductive capacity of medium. Download India's Leading JEE | NEET | Class 9,10 Exam preparation app, Vector form of Coulomb’s Law – Definition | Examples. Three equal charges Q each are placed on the vertices of an equilateral triangle of side a. Force is a vector quantity as it has both magnitude and direction. Two equally charged identical metal spheres A and B repel each other with a force of 2 × 10, Force F is acting between two charges. {{{{(1.6 \times {{10}^{ – 19}})}^2}} \over {{r^2}}}$N. Sol. In applying it to the case of extended bodies of finite size care should be taken in assuming the whole charge of a body to be concentrated at its ‘center’ as this is true only for spherically charged body, that too for a external point. $F = {1 \over {4\pi {\varepsilon _0}}}{{q\left( {Q – q} \right)} \over {{r^2}}}$ The dielectric constants of different mediums are: $\mathop {{F_{21}}}\limits^ \to $= force on q2 due to q1, $\mathop {{F_{21}}}\limits^ \to = {1 \over {4\pi {\varepsilon _0}{\varepsilon _r}}}\,\,\,{{{q_1}{q_2}} \over {r_{12}^2}}\,\,{\hat r_{12}}$, $\mathop {{F_{12}}}\limits^ \to $= Force on q1 due to q2, $\mathop {{F_{12}}}\limits^ \to = {1 \over {4\pi {\varepsilon _0}{\varepsilon _r}}}\,\,\,{{{q_1}{q_2}} \over {r_{21}^2}}\,\,{\hat r_{21}}$, $\mathop {{F_{12}}}\limits^ \to = – \mathop {{F_{21}}}\limits^ \to $ (∵ ${\hat r_{12}} = – {\hat r_{21}}$ ), Or ${\mathop F\limits^ \to _{12}} + {\mathop F\limits^ \to _{21}} = 0$. We denote force on q1 due to q2 by F12 and Write down Coulomb’s law in vector form and mention what each term represents. A French physicist Charles Augustin de Coulomb in 1785 coined a tangible relationship in mathematical form between two bodies that have been electrically charged. The electrical force, like all forces, is typically expressed using the unit Newton. From symmetry the direction is along y-axis. The electrostatic force is an action-reaction pair, i.e., the force exerted by one charge on the other is equal and opposite to the force exerted by the other on the first. So, the net force on C due to charges on A and B, The vector form of the Coulomb’s law is independent of the nature of the sign carried by the charges because of the fact that both the forces are opposite in nature. which depends upon system of units and also on medium between two charges. How should we divide a charge ‘Q’ to get maximum force of repulsion between them? What is the net electric force on C? With F1 = F2 = kQ2/a2 $F = {{\sqrt 3 k{Q^2}} \over {{a^2}}}$ Dimensions of permittivity, ${\varepsilon _0} = {{{Q^2}} \over {F\,\, \times lengt{h^2}}}$$ = {{{T^2}{A^2}} \over {ML{T^{ – 2}}{L^2}}}$= M–1 L–3 T4A2. $F = {1 \over {4\pi {\varepsilon _0}}}{{q \times q} \over {{r^2}}} = 2 \times {10^{ – 5}}N$, When sphere C touches B, the charge of B, q will distribute equally on B and C as spheres are identical now: qB = qC = (q/2), ${F_{CA}} = {1 \over {4\pi {\varepsilon _0}}}{{q\left( {q/2} \right)} \over {{{\left( {r/2} \right)}^2}}} = 2F$ along $\mathop {AB}\limits^ \to $ due to charge on A, and, ${F_{CB}} = {1 \over {4\pi {\varepsilon _0}}}{{(q/2)\,(q/2)} \over {{{(r/2)}^2}}} = F$ along $\mathop {BA}\limits^ \to $ due to charge on B. Derive an expression of the relation vector J = σ E (equivalent form of Ohm’s law). Create papers in minutes Print with your name & Logo Download as PDF 3 Lakhs+ Questions Now, since both the charges are of the same sign, there will be a repulsive force between t… Thus if two point charges q1 and q2 are separated by a distance r in vacuum, the magnitude of the force F between them is given by, The constant k in eqn. The direction of the electrical force is dependent upon whether the charged objects are charged with like charge or opposite charge. or $F’ = {1 \over {4\pi { \in _0}K}}{{{q_1}{q_2}} \over {{r^2}}}$ If q1 & q2 are charges, r is the distance between them and F is the force acting between them. Write Coulomb’s law in vector notation and explain it. The electrostatic force is a medium dependent force. We denote force on q1 due to q2 by F12 and force on q2 by q1 by F21 as shown in figure. The resultant force $F = \sqrt {F_1^2 + F_2^2 + 2{F_1}{F_2}\cos 60^\circ } $ $F = {1 \over {4\pi {\varepsilon _0}}}{{q\left( {Q – q} \right)} \over {{r^2}}}$, or ${1 \over {4\pi {\varepsilon _0}}}{{Q – 2q} \over {{r^2}}} = 0$. If q1 & q2are charges, r is the distance between them and F is the force acting between them Then, F ∝ q1q2 F ∝ 1/r² ∴ F ∝ q1q2r2q1q2r2 Or F=Cq1q2r2F=Cq1q2r2 C is const. This video explains the basic properties of electric charges. What is the resultant force on any one charge due to the other two? Q. (b) Both the forces obey inverse square law i.e., F ∝ (1/r²). or ${1 \over {4\pi {\varepsilon _0}}}{{Q – 2q} \over {{r^2}}} = 0$ For an electron-proton system, electrostatic force of attraction, ${F_e} = {1 \over {4\pi {\varepsilon _0}}}{{e.e} \over {{r^2}}} = 9 \times {10^9}. (vector quantity is given in Bold) Let the position vectors of charges q1 and q2 be r1 and r2 respectively. Want a call from us give your mobile number below, For any content/service related issues please contact on this number. Coulomb measured the force between two point charges and found that it varied inversly as the square of the distance between the charges and was directly proportional to the product of the magnitude of two charges and acted along the line of joining the two charges. Coulomb measured the force between two point charges and found that it varied inversly, as the square of the distance between the charges and was directly proportional to the product of the. No need to write separate equation for like and unlike charges. C is const. ${F_{CA}} = {1 \over {4\pi {\varepsilon _0}}}{{q\left( {q/2} \right)} \over {{{\left( {r/2} \right)}^2}}} = 2F$ along $\mathop {AB}\limits^ \to $ due to charge on A SHOW SOLUTION (c) Electrostatic forces are very large as compared to gravitational forces. The charges are shown in fig. Since force is vector, we need to write Coulombs law in vector notation. When sphere C touches B, the charge of B, q will distribute equally on B and C as spheres are identical now: qB = qC = (q/2) SHOW SOLUTION The above equation is valid for any sign of q1 and q2. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students. (c) Both are conservative forces, i.e., the work done by these is independent of the nature of path. This force acts along the line joining the center of two charges. Derive the expression Coulomb's Law in Vector form.

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