Chapter 23: Electric Fields - Comprehensive Notes

Electric Fields

Charge

Charge is represented by the symbols $Q$ or $q$ .
There are two types of charges: positive ( $+$ ) and negative (-$).
Like charges repel each other, while opposite charges attract each other.
Point charges are idealized charges with no dimensions.
The elementary charge, which is the magnitude of the charge of an electron or proton, is given by: e = 1.6
imes 10^{-19} C $</li><li>To charge an object negatively, electrons must be added.</li></ul><h4 id="7a0e3a7c-f7de-48fd-a7a8-c30d3f17bdc5" data-toc-id="7a0e3a7c-f7de-48fd-a7a8-c30d3f17bdc5" collapsed="false" seolevelmigrated="true">Coulomb’s Law</h4><ul><li>Coulomb's Law describes the electric force between two point charges:$ F = k frac{q1 q2}{r^2} $where:<ul><li>$ F $is the electric force.</li><li>$ q1 $and$ q2 $are the magnitudes of the charges.</li><li>$ r $is the distance between the charges.</li><li>$ k $is Coulomb's constant.</li></ul></li><li>Coulomb's constant$ k $is given by: $ k = 9
imes 10^9 Nm^2/C^2 $</li><li>$ k $can also be expressed in terms of the permittivity of free space$ \&epsilono $:$ k = \frac{1}{4 \pi \varepsilono} $ where the permittivity of free space is: $ \varepsilon_o = 8.85
imes 10^{-12} C^2/Nm^2 $</li><li>The direction of the force is along the line connecting the two charges; attractive if the charges are of opposite signs and repulsive if the charges are of the same sign.</li></ul><h4 id="e107b5aa-ecbc-4f32-8e62-9cad31973e83" data-toc-id="e107b5aa-ecbc-4f32-8e62-9cad31973e83" collapsed="false" seolevelmigrated="true">Example 23.1: Electric Force between Electron and Proton in Hydrogen Atom</h4><ul><li>Problem: Find the electric force between the electron and proton in a hydrogen atom, where the separation distance is$ 5.3
imes 10^{-11} m $.</li><li>Solution:<ol><li>Apply Coulomb's Law:$ F = k \frac{qe qp}{r^2} $</li><li>Substitute the values: $ F = (9
imes 10^9 Nm^2/C^2) \frac{(1.6
imes 10^{-19} C)(1.6
imes 10^{-19} C)}{(5.3
imes 10^{-11} m)^2} $</li><li>Calculate the force: $ F = 8.2
imes 10^{-8} N $</li></ol></li></ul><h4 id="f214431b-8414-4337-8cee-399b3daaed99" data-toc-id="f214431b-8414-4337-8cee-399b3daaed99" collapsed="false" seolevelmigrated="true">Problem 23.7: Resultant Electric Force on a Charge in an Equilateral Triangle</h4><ul><li>Problem: Three point charges are located at the corners of an equilateral triangle. Calculate the resultant electric force on the$ 7 \mu C $charge.</li><li>Given: Equilateral triangle with side length$ 0.5 m $. Two forces,$ F{13} $and$ F{23} $, act on the$ 7 \mu C $charge.</li><li>Calculations:<ul><li>Calculate the magnitudes of the forces$ F{13} $and$ F{23} $: $ F{13} = k \frac{q1 q3}{r^2} = (9 imes 10^9) \frac{(7 imes 10^{-6})(2 imes 10^{-6})}{(0.5)^2} = 0.5 NF{23} = k \frac{q2 q3}{r^2} = (9
imes 10^9) \frac{(7
imes 10^{-6})(4
imes 10^{-6})}{(0.5)^2} = 1 N $</li><li>Resolve the forces into components:<ul><li>$ F{13} $components:$ F{13x} = F{13} \cos(60^\circ) = 0.5 \cos(60^\circ) = 0.25 NF{13y} = F_{13} \sin(60^\circ) = 0.5 \sin(60^\circ) = 0.43 N $</li><li>$ F{23} $components:$ F{23x} = -F{23} \cos(60^\circ) = -1 \cos(60^\circ) = -0.5 NF{23y} = -F_{23} \sin(60^\circ) = -1 \sin(60^\circ) = -0.86 N $</li></ul></li><li>Calculate the resultant force components: $ Fx = F{13x} + F{23x} = 0.25 N - 0.5 N = -0.25 NFy = F{13y} + F{23y} = 0.43 N - 0.86 N = -0.43 N $</li><li>Calculate the magnitude of the resultant force: $ F = \sqrt{Fx^2 + Fy^2} = \sqrt{(-0.25)^2 + (-0.43)^2} = 0.5 N $</li><li>Calculate the direction of the resultant force: $ \theta = \arctan \frac{Fy}{Fx} = \arctan \frac{-0.43}{-0.25} = 59.8 \approx 240 $</li></ul></li></ul><h4 id="78a306b7-3e09-48de-b479-9d4410764dce" data-toc-id="78a306b7-3e09-48de-b479-9d4410764dce" collapsed="false" seolevelmigrated="true">Example 23.3: Position for Zero Resultant Force</h4><ul><li>Problem: Three point charges lie along the x-axis. Find the x-coordinate of$ q_3 $such that the resultant force on it is zero.</li><li>Given: Charges$ q1 $,$ q2 $, and$ q_3 $on the x-axis.</li><li>Solution:<ol><li>Set the magnitudes of the forces equal:$ F{13} = F{23} $</li><li>Apply Coulomb's Law:$ k \frac{q1 q3}{x^2} = k \frac{q2 q3}{(x_2 - x)^2} $</li><li>Simplify:$ \frac{q1}{x^2} = \frac{q2}{(x_2 - x)^2} $</li><li>Substitute values (e.g.,$ q1 = 15 \rimes 10^{-6} C $,$ q2 = 6 \rimes 10^{-6} C $): $ \frac{15 \rimes 10^{-6}}{x^2} = \frac{6 \rimes 10^{-6}}{(2 - x)^2} $</li><li>Solve for$ x $: $ 15(2 - x)^2 = 6x^2 $ $ 15(4 - 4x + x^2) = 6x^2 $ $ 60 - 60x + 15x^2 = 6x^2 $ $ 9x^2 - 60x + 60 = 0 $ $ 3x^2 - 20x + 20 = 0 $</li><li>Use the quadratic formula to find the roots. Given the general quadratic equation$ ax^2 + bx + c = 0 $, the solution is given by $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ Using this formula, we have $ x = \frac{20 \pm \sqrt{(-20)^2 - 4(3)(20)}}{2(3)} $ $ x = \frac{20 \pm \sqrt{400 - 240}}{6} $ $ x = \frac{20 \pm \sqrt{160}}{6} $ $ x = \frac{20 \pm 4\sqrt{10}}{6} $ $ x = \frac{10 \pm 2\sqrt{10}}{3} $</li><li>Obtain the two possible coordinates. $ x = \frac{10 + 2\sqrt{10}}{3} = 4.27 m $ $ x = \frac{10 - 2\sqrt{10}}{3} = 2.4 m $</li></ol></li></ul><h4 id="4e2d21ad-36e0-4ca0-ac18-9bd3d8a3294b" data-toc-id="4e2d21ad-36e0-4ca0-ac18-9bd3d8a3294b" collapsed="false" seolevelmigrated="true">Exam Question: Electric Force on a Point Charge</h4><ul><li>Problem: Two point charges$ +Q $and$ -Q $are placed as shown. Find the magnitude of the electric force on a point charge$ +q $located at the origin.</li><li>Possible Answers: A)$ \frac{kQq}{d^2} $ B)$ \frac{2kQq}{d^2} $ C)$ \frac{\sqrt{2}kQq}{d^2} $ D)$ \frac{2\sqrt{2}kQq}{d^2} $</li></ul><h4 id="fc55be0f-8a7f-4a87-ad59-9fc37726a68c" data-toc-id="fc55be0f-8a7f-4a87-ad59-9fc37726a68c" collapsed="false" seolevelmigrated="true">Electric Field</h4><ul><li>The electric field of a point charge is given by: $ E = k \frac{q}{r^2} $</li><li>The unit of electric field is Newtons per Coulomb (N/C).</li><li>The relationship between electric force and electric field is: $ F = qE $</li></ul><h4 id="bdb6fd1f-fdcb-4815-ab4d-ba1175156979" data-toc-id="bdb6fd1f-fdcb-4815-ab4d-ba1175156979" collapsed="false" seolevelmigrated="true">Example 23.5: Electric Field at a Point</h4><ul><li>(A) Find the electric field at point P due to a charge. Given$ q = 7 \rimes 10^{-6} C $and$ r = 0.4 m $<ol><li>$ E = k \frac{q}{r^2} = (9 \rimes 10^9) \frac{7 \rimes 10^{-6}}{(0.4)^2} = 3.9 \rimes 10^5 N/C $</li><li>The electric field vector is:$ \vec{E} = 3.9 \rimes 10^5 \hat{j} N/C $</li></ol></li><li>(B) Find the electric field at point P due to two charges. Given$ q1 = 7 \rimes 10^{-6} C $at a distance of$ 0.4 m $and$ q2 = 5 \rimes 10^{-6} C $at a distance of$ 0.5 m $.<ol><li>Calculate$ E1 $:$ E1 = k \frac{q1}{r1^2} = (9 \rimes 10^9) \frac{7 \rimes 10^{-6}}{(0.4)^2} = 3.9 \rimes 10^5 N/C $</li><li>Calculate$ E2 $:$ E2 = k \frac{q2}{r2^2} = (9 \rimes 10^9) \frac{5 \rimes 10^{-6}}{(0.5)^2} = 1.8 \rimes 10^5 N/C $</li><li>Find the angle$ \theta $:$ \theta = \arctan(\frac{0.4}{0.3}) = 53^\circ $</li><li>Resolve$ E2 $into components:$ E{2x} = E2 \cos(53^\circ) = 1.8 \rimes 10^5 \cos(53^\circ) = 1.1 \rimes 10^5 N/CE{2y} = -E_2 \sin(53^\circ) = -1.8 \rimes 10^5 \sin(53^\circ) = -1.4 \rimes 10^5 N/C $</li><li>Calculate the resultant electric field components: $ Ex = E{1x} + E{2x} = 0 + 1.1 \rimes 10^5 = 1.1 \rimes 10^5 N/CEy = E{1y} + E{2y} = 3.9 \rimes 10^5 - 1.4 \rimes 10^5 = 2.5 \rimes 10^5 N/C $</li><li>The resultant electric field is: $ \vec{E} = 1.1 \rimes 10^5 \hat{i} + 2.5 \rimes 10^5 \hat{j} N/C $</li><li>Calculate the magnitude of the resultant electric field: $ E = \sqrt{Ex^2 + Ey^2} = \sqrt{(1.1 \rimes 10^5)^2 + (2.5 \rimes 10^5)^2} = 2.7 \rimes 10^5 N/C $</li><li>Find the direction of the resultant electric field: $ \phi = \arctan \frac{2.5 \rimes 10^5}{1.1 \rimes 10^5} = 66^\circ $</li></ol></li></ul><h4 id="3cfe1fa0-7dda-405c-b4db-5e0df2d529e1" data-toc-id="3cfe1fa0-7dda-405c-b4db-5e0df2d529e1" collapsed="false" seolevelmigrated="true">Problem 23.15: Electric Field is Zero at Point P</h4><ul><li>Problem: Find the value of$ x $where the electric field is zero at point P.</li><li>Solution:<ol><li>Set the electric fields equal:$ E1 = E2 $</li><li>$ k \frac{q1}{x^2} = k \frac{q2}{(1 + x)^2} $</li><li>$ \frac{2.5 \rimes 10^{-6}}{x^2} = \frac{6 \rimes 10^{-6}}{(1 + x)^2} $</li><li>Solve for$ x $: $ 2.5(1 + x)^2 = 6x^2 $ $ 2.5(1 + 2x + x^2) = 6x^2 $ $ 2.5 + 5x + 2.5x^2 = 6x^2 $ $ 3.5x^2 - 5x - 2.5 = 0 $ $ 7x^2 - 10x - 5 = 0 $</li><li>Use the quadratic formula to solve for x: $ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(7)(-5)}}{2(7)} $ $ x = \frac{10 \pm \sqrt{100 + 140}}{14} $ $ x = \frac{10 \pm \sqrt{240}}{14} $</li></ol></li></ul>$ x = \frac{10 \pm 4\sqrt{15}}{14} $$ x = \frac{5 \pm 2\sqrt{15}}{7} $ 6. Solve the two possible cooridnates of x $ x = 1.82 m $ $ x = -0.39 m $<h4 id="0a85f054-7ff6-4762-a5f6-c2941d56ffb0" data-toc-id="0a85f054-7ff6-4762-a5f6-c2941d56ffb0" collapsed="false" seolevelmigrated="true">Electric Field of Continuous Charge Distribution</h4><ul><li>Linear charge density ($ \lambda $):$ \lambda = \frac{dq}{dl} $</li><li>Surface charge density ($ \sigma $):$ \sigma = \frac{dq}{dA} $</li><li>Volume charge density ($ \rho $):$ \rho = \frac{dq}{dV} $</li><li>The electric field due to a continuous charge distribution is calculated by integrating over the distribution: $ E = \int k \frac{dq}{r^2} $</li></ul><h4 id="d6f0d332-eed7-485e-aed5-7a2ebfe6dcce" data-toc-id="d6f0d332-eed7-485e-aed5-7a2ebfe6dcce" collapsed="false" seolevelmigrated="true">Example 23.7: Electric Field of a Charged Rod</h4><ul><li>Problem: A rod of length$ \lambda $has a uniform charge density$ \lambda $. Find the electric field at point P located along the axis of the rod, a distance$ a $from one end.</li><li>Solution:<ol><li>$ dE = k \frac{dq}{x^2} $</li><li>Since$ dq = \lambda dx $: $ dE = k \frac{\lambda dx}{x^2} $</li><li>Integrate from$ a $to$ a + \lambda $: $ E = \int{a}^{a+\lambda} k \frac{\lambda}{x^2} dxE = k \lambda \int{a}^{a+\lambda} \frac{1}{x^2} dx $ $ E = k \lambda [-\frac{1}{x}]_{a}^{a+\lambda} $ $ E = k \lambda (\frac{1}{a} - \frac{1}{a+\lambda}) $ $ E = k \lambda (\frac{a+\lambda - a}{a(a+\lambda)}) $ $ E = \frac{k \lambda^2}{a(a+\lambda)} $</li></ol>$ \vec{E} = \frac{k \lambda^2}{a(a+\lambda)} \hat{i} $</li><li>Electric force on a point charge$ q $at point P: $ F = qE $</li></ul>$ \vec{F} = \frac{kq \lambda^2}{a(a+\lambda)} \hat{i} $<h4 id="7e760c48-13db-42cf-a9cc-a1da505e9ae4" data-toc-id="7e760c48-13db-42cf-a9cc-a1da505e9ae4" collapsed="false" seolevelmigrated="true">Example 23.8: Electric Field of a Charged Ring</h4><ul><li>Problem: A ring of radius$ a $carries a uniformly distributed positive total charge$ Q $. Find the electric field at a point P lying a distance$ x $from its center along the central axis perpendicular to the plane of the ring.</li><li>Solution:<ol><li>$ dE = k \frac{dq}{r^2} \cos(\theta) $</li><li>$ E = \int k \frac{dq}{r^2} \cos(\theta) $</li><li>$ E = k \frac{Q}{r^2} \cos(\theta) $</li><li>Since$ \cos(\theta) = \frac{x}{r} $and$ r = \sqrt{x^2 + a^2} $: $ E = \frac{kQx}{(\sqrt{x^2+a^2})^3} $</li></ol>$ E = \frac{kQx}{(x^2+a^2)^{3/2}} $</li><li>At the center of the ring (x=0), the electric field is zero.</li></ul><h4 id="fec19ed3-33bb-491d-afd0-d430d5cfebb7" data-toc-id="fec19ed3-33bb-491d-afd0-d430d5cfebb7" collapsed="false" seolevelmigrated="true">Electric Field Lines</h4><ul><li>The electric field vector is tangent to the electric field line at each point.</li><li>The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region.</li></ul><h4 id="477f1049-f51c-4a33-9937-719df339196c" data-toc-id="477f1049-f51c-4a33-9937-719df339196c" collapsed="false" seolevelmigrated="true">Problem 23.40: Ratio of Charges</h4><ul><li>Problem: Determine the ratio$ \frac{q1}{q2} $.</li><li>Given:$ 6q1 = -18q2 $</li><li>Solution: $ \frac{q1}{q2} = \frac{-18}{6} = -3 $</li></ul><h4 id="061aa894-d9c9-4e85-9b14-21e8611a19fa" data-toc-id="061aa894-d9c9-4e85-9b14-21e8611a19fa" collapsed="false" seolevelmigrated="true">Motion of a Point Charge in a Uniform Electric Field</h4><ul><li>A uniform electric field$ \vec{E} $gives a constant acceleration: $ \sum F = ma \rightarrow qE = ma \rightarrow a = \frac{qE}{m} $</li><li>Kinematic equations:</li></ul>$ \vec{xf} = \vec{xi} + \vec{v_i} t + \frac{1}{2} \vec{a} t^2 $$ {vf}^2 = {vi}^2 + 2 a \Delta x $<h4 id="2df0f8f2-bb1f-4ea9-9b7d-af2ce6929752" data-toc-id="2df0f8f2-bb1f-4ea9-9b7d-af2ce6929752" collapsed="false" seolevelmigrated="true">Example 23.10: Speed of a Charge in a Uniform Electric Field</h4><ul><li>Problem: A positive point charge$ q $of mass$ m $is released from rest at point A in a uniform electric field$ E $. Find its speed at point B.</li><li>Given:$ v_i = 0 $,$ \Delta x = d $,$ a = \frac{qE}{m} $</li><li>Solution:</li></ul>$ {vf}^2 = {vi}^2 + 2 a \Delta x $$ {v_f}^2 = 0 + 2 \frac{qE}{m} d $$ {v_f} = \sqrt{\frac{2qEd}{m}}$$