{"id":3896,"date":"2023-10-06T06:24:17","date_gmt":"2023-10-06T06:24:17","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3896"},"modified":"2023-10-06T06:27:32","modified_gmt":"2023-10-06T06:27:32","slug":"week-3-ss3-first-term-physics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-3-ss3-first-term-physics-notes\/","title":{"rendered":"Week 3 &#8211; SS3 First Term Physics Notes"},"content":{"rendered":"<p><strong>WEEK 3<br \/>\n<\/strong><strong>Electric Field<br \/>\n<\/strong><strong>CONTENTS<br \/>\n<\/strong><\/p>\n<ul>\n<li>\n<div>Electric Field\n<\/div>\n<\/li>\n<li>\n<div>Coulomb&#8217;s Law\n<\/div>\n<\/li>\n<li>\n<div>Electric Field Intensity\n<\/div>\n<\/li>\n<li>\n<div>Electric Potential\n<\/div>\n<\/li>\n<\/ul>\n<p><strong>ELECTRIC FIELD<\/strong><br \/>\n\t\tAn electric field is a region of space which surrounds a system of electric charges. Electrical forces will act on any electric charge which is placed within the region.  Electric field is a vector quantity.  The direction of the filed can be determined using a test charge (a small positive charge)<br \/>\nFundamental Law of Electrostatics<br \/>\nThe fundamental law of electrostatic states that: <em>&#8220;Like charge repels, unlike charges attract.<\/em><br \/>\n\t\t<strong>COULOMB&#8217;S LAW<br \/>\n<\/strong>Coulomb&#8217;s law states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.<br \/>\nMathematically,<br \/>\nF \u03b1 q<sub>1<\/sub>q<sub>2<\/sub><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0r<sup>2<br \/>\n<\/sup>F =k q<sub>1<\/sub>q<sub>2<\/sub><br \/>\n\t\t\t\u00a0\u00a0\u00a0\u00a0r<sup>2<br \/>\n<\/sup>Where k = 1 = 8.99&#215;10<sup>9<\/sup><br \/>\n\t\t     4\u03c0\u025b<sub>\u25e6<\/sub><br \/>\n\t\tThus,<br \/>\nF = q<sub>1<\/sub>q<sub>2<\/sub><br \/>\n\t\t\t      4\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2<\/sup><br \/>\n\t\t\t<strong>ELECTRIC FIELD INTENSITY OR STRENGTH (E<\/strong>)<br \/>\nThe electric field intensity, E, at any point in an electric field is the force experienced by a unit positive test charge at that point.  It is a vector quantity whose S. I unit is (N\/C), mathematically.<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0E = F<br \/>\n\t\t       Q<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi1.png\" alt=\"\"\/>E= Electric field intensity (NC<sup>-1<\/sup>); F = Force, q = charge.<br \/>\nr\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0Q\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0q<br \/>\nFrom the diagram above, F between Q and q is given as<br \/>\nF = Qq<br \/>\n      4\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2<\/sup><br \/>\n\t\t\tBut E = F         = \u00a0\u00a0\u00a0\u00a0Qq x 1<br \/>\n        Q\u00a0\u00a0\u00a0\u00a04\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2\u00a0\u00a0\u00a0\u00a0 <\/sup>q<br \/>\n  :. E = Q<br \/>\n\t\t4\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2<\/sup><br \/>\n\t\t<strong>ELECTRIC POTENTIAL<br \/>\n<\/strong>The electric potential (V) at a point is the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.  It is measured in volts.  It is scalar quantity.<br \/>\nMathematically, v = w<br \/>\n    q<br \/>\n\t\t\tWhere V= electric potential (volts); W= work done in joules; q = charge in coulombs<br \/>\nThe electric potential at a point due to a charge Q at a distance r from the charge Q at a distance r from the charge is given as:<br \/>\nV = Q<br \/>\n     4\u03c0\u025b<sub>\u25e6<\/sub>r<br \/>\nIf the work done is against the field, the potential is positive.  If the work done is by the field, the potential is negative.  The potential an infinity is zero.  Also the potential of the earth is zero.  The earth is used to test the potential of the body.  This is done by connecting a wire form the body to the earth (the body is said to be earthed).  If electrons flow from the body to the earth, the body is at a negative potential.  If electron flows from the earth to the body, the body is at positive potential.  Positive points are at higher potential while negative points are at lower potential.<br \/>\n<strong>POTENTIAL DIFFERENCE<\/strong><br \/>\n\t\tThe potential difference between any two points in an electric field is the work done is taking a unit positive charge from one point to another in the field.<br \/>\nIf a charge Q is moved from a point at a potential V<sub>1<\/sub> to another at a potential V<sub>2<\/sub>, the potential difference (V<sub>1<\/sub> \u2013 V<sub>2<\/sub>) is the work done by the field.<br \/>\nWork done on the charge, W = Q (V<sup>1<\/sup> \u2013 V<sup>2<\/sup>)<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi2.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0         Q<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  A\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0    B<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x<br \/>\nIf Q moves from A to B, then the work done,<br \/>\nW = Force x distance<br \/>\nW = F.X<br \/>\nBut E = F<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0   Q<br \/>\nEQ = F<br \/>\n: .W = EQ.X<br \/>\nBut W = Q(Va \u2013 Vb)<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi3.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0        Q (Va \u2013 Vb) = EQ. X<br \/>\nVa \u2013 Vb = E X<br \/>\nE =  Va \u2013 Vb<br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0 X<br \/>\n: .  E =p.d.<br \/>\n\t\tdistance<br \/>\ni.e, E = V<br \/>\n\t\t   X<br \/>\nV = E X<br \/>\n: .  V =Q<br \/>\n\t\t4\u03c0\u025b<sub>\u25e6<\/sub><\/p>\n<p>\u00a0ELECTRON VOLT (eV)<br \/>\nThe electron volt is the quantity of energy gained by an electron in accelerating through a potential difference of one volt.<br \/>\nElectronic charge = 1.6 x 10<sup>-1<\/sup>9C<br \/>\nI e V = 1.6 x 10<sup>-19<\/sup> x 1 = 1.6 x 10<sup>-19<\/sup>J<br \/>\nThe energy acquired by a charged particle accelerated by an electronic field in a vacuum depends only on its charge and the p.d. through which it falls.  When the electron is in motion, its kinetic energy will be \u00bd mv<sup>2<\/sup>.  If the electron moves in a circle of radius r, the force towards the centre in mv<sup>2<\/sup> (centripetal force), and it is provided by the electrical force of attraction<br \/>\nForce of attraction   =     e<sup>2<\/sup><br \/>\n\t\t4\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2<\/sup><br \/>\n\t\t: . \u00bd mv<sup>2<\/sup> =     e<sup>2<\/sup><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi4.png\" alt=\"\"\/>4\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2<\/sup><br \/>\n\t\t=\u00a0\u00a0\u00a0\u00a0<sup>1<\/sup>\/<sub>2<\/sub><br \/>\n\t\t\te<sup>2<\/sup><br \/>\n\t\t4\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0<strong><br \/>\n\t\t\t<\/strong>WORKED EXAMPLE<br \/>\n1.\u00a0\u00a0\u00a0\u00a0Calculate the energy in eV and in Joule of an \u03b1 particle (helium nucleus) accelerated through a p.d. of 4 x 10<sup>6<\/sup>V.<br \/>\nSOLUTION<strong><br \/>\n\t\t\t<\/strong>The charge on an \u03b1 particle is 2e.<br \/>\nKE = work done<br \/>\nKE = charge x p.d. = q x v<br \/>\n= 2 x 4 x10<sup>6<\/sup><br \/>\n\t\t= 8 x 10<sup>6<\/sup> eV = 8 MeV<br \/>\n1eV = 1.6 x 10<sup>-19<\/sup>J.<br \/>\nKE gained = 8 x 10<sup>6<\/sup> x 1.6 x 10<sup>-19<\/sup><br \/>\n\t\t= 1.48 x 10<sup>-12<\/sup>J<br \/>\n2.\u00a0\u00a0\u00a0\u00a0An electron gun releases an electron.  The p.d. between the gun and the collector plate is 100V.  What is the velocity of the electron just before it touches the collector plate?  (e = -1.6 x 10<sup>-19<\/sup>C, M<sub>e<\/sub> = 9.1 x 10<sup>-31<\/sup> kg)<br \/>\nSOLUTION<strong><br \/>\n\t\t\t<\/strong>Kinetic energy = QV<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0   = 100 x 1.6 x 10<sup>-19<\/sup><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0   = 1.6 x 10<sup>-19<\/sup>J<br \/>\nAlso, kinetic energy = \u00bdMV<sup>2<br \/>\n<\/sup>Thus, \u00bdM<sub>e<\/sub>V<sup>2<\/sup> = 1.6 x 10<sup>-19<\/sup>J<br \/>\n\u00bd (9.1 x 10<sup>-31<\/sup>) V<sup>2<\/sup> = 1.6 x 10<sup>-19<\/sup><\/p>\n<p>\u00a0V<sup>2<\/sup> = 3.2 x 10<sup>-16<br \/>\n<\/sup>9.1 x 10<sup>-31<br \/>\n<\/sup>V<sup>2<\/sup> = 0.35 x 10<sup>14<br \/>\n<\/sup>: .V = 6 x 10<sup>6 <\/sup>ms<sup>-1<\/sup><br \/>\n\t\t<strong>CAPACITORS AND CAPACITANCE<br \/>\n<\/strong><strong>CAPACITORS<br \/>\n<\/strong>A Capacitor is a device for storing electrical energy or charges. In general, capacitors can be in the form of two conductors which are insulated electrically from the surroundings. However, most common types of capacitors are in the form of two parallel plate conductors which are separated by a very small distance, d. The two plates of the capacitor can be made to carry equal and opposite charges by connecting the capacitor across the terminals of a battery such that the potential difference across the plate is V.<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi5.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi6.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi7.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi8.png\" alt=\"\"\/>Capacitor is represented as<br \/>\n<strong>CAPACITANCE<br \/>\n<\/strong>The capacitance of a capacitor is defined as the ratio of the charge Q on either conductor to the potential difference V between the two conductors<br \/>\nC = Q\/V<br \/>\nQ = CV<br \/>\nThe SI unit of capacitance is the farad (F) which is equivalent to coulomb per volts (CV<sup>-1<\/sup>)<br \/>\nFactors that affect the capacitance of a capacitor are:<\/p>\n<ol>\n<li>\n<div>The area of the plates\n<\/div>\n<\/li>\n<li>\n<div>The separation between the plates\n<\/div>\n<\/li>\n<li>\n<div>The di-electric substance between the plates\n<\/div>\n<\/li>\n<\/ol>\n<p>For a parallel plate capacitor, the capacitance C is given by<br \/>\nC =\u025bA<br \/>\n\u00a0\u00a0\u00a0\u00a0  d<br \/>\nWhere:<br \/>\nA= area of the plates<br \/>\nd= their separation<br \/>\n\u025b= permittivity of the dielectric medium (Fm<sup>-1<\/sup>)<br \/>\n<strong>CAPACITOR IN SERIES AND IN PARALLEL<\/strong><sup><br \/>\n\t\t\t<\/sup>If two or more capacitors is c<sub>1<\/sub>, c<sub>2 \u2026 <\/sub>are connected in series, it can be shown that the equivalent or net capacitance, c of the combination is given by:<br \/>\n1\/c = 1\/c<sub>1<\/sub> + 1\/c<sub>2<\/sub> +\u2026<br \/>\nIf they are connected in parallel the net capacitance C in this is given by:<br \/>\nC = c<sub>1<\/sub> + c<sub>2<\/sub> + \u2026<br \/>\nNote that the opposite is the case if these were resistance.<br \/>\n<strong>SIMPLE PROBLEMS<br \/>\n<\/strong>A capacitor contain a charge of 4 .0 x 10<sup>\u2013 4 <\/sup>coulomb when a potential difference of 400 v is applied across it. Calculate the capacitance of the capacitor<br \/>\nThe capacitance C = q\/v<br \/>\n= 4.0 x 10<strong><sup>-4<\/sup><\/strong><sup><br \/>\n\t\t\t<\/sup>400<br \/>\n= 10<sup> \u2013 6<\/sup>F<sup><br \/>\n\t\t\t<\/sup>= 1.0<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi9.png\" alt=\"\"\/>f<\/p>\n<p>\u00a0<\/p>\n<h2>ENERGY STORED IN CAPACITOR<br \/>\n<\/h2>\n<p>A charged is a store of electrical energy. When a charge, q , is moved through a p.d , the work done is given by<br \/>\n       W = average p.d x charge<br \/>\n          = \u00bd qv = \u00bd QV<br \/>\n      But v = q \/c; V = Q<br \/>\n     C<br \/>\nW= \u00bd <sup>q<\/sup>\/c x q = \u00bd q<sup>2<\/sup>\/c<br \/>\n<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi10.png\" alt=\"\"\/>W = \u00bd <sup>Q<\/sup>\/C<br \/>\nUsing Q=CV<br \/>\nW =1\/2CV<sup>2<\/sup><br \/>\n\t\tTherefore the work done is either<br \/>\n\u00bc q<sup>2<\/sup>\/c or <sup>1<\/sup>\/2cv<sup>2\u00a0\u00a0\u00a0\u00a0<img decoding=\"async\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi11.png\" alt=\"\"\/><\/sup>W = \u00bd CV<sup>2<\/sup><br \/>\n\t\tThis work is stored in the capacitor as electrical potential energy<strong><br \/>\n\t\t\t<\/strong><strong>CLASSWORK<br \/>\n<\/strong>1. (a)<strong><br \/>\n\t\t\t<\/strong>State Coulomb&#8217;s law (b) Calculate the electric field intensity in vacuum at a distance of 5cm from a charge of 5.0x 10<sup>-4<\/sup>c (1\/4\u03c0\u025b<sub>\u25e6<\/sub> = 9.0 x 10<sup>9<\/sup> NM<sup>2<\/sup>C<sup>-2<\/sup>).<strong><br \/>\n\t\t\t<\/strong>2 (a) Define electric field intensity (b) Two similar but opposite point charges \u2013q and +q each of magnitude 6\u00b5C are separated by a distance of 12cm in vacuum as shown below:<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi12.png\" alt=\"\"\/><br \/>\n\t\tr\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0+q\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-q<br \/>\nCalculate the magnitude and direction of the resultant electric field intensity at p<br \/>\n3. Calculate the electric potential due to a positive charge of 10-<sup>12<\/sup>C at a point distance 10cm away (   1\/4\u03c0 = 9.0 x 10<sup>9<\/sup>m\/F)<br \/>\n4. (a) Explain the term capacitor (b) Give three factors that can affect the capacitance of a capacitor (c) The net charge on capacitor which is charged to a p.d of 200 is 1.0 x 10<sup>-4<\/sup> coulomb. What is the capacitance of capacitor and the energy stored in the capacitor?<br \/>\n5.  A point, A, is a potential of 120v.  Determine the work done in moving an electric charge 25C from A to B.<br \/>\n<strong>ASSIGNMENT<br \/>\n<\/strong><strong>SECTION A<br \/>\n<\/strong><\/p>\n<ol>\n<li>\n<div>The electric force between two-point charges each of magnitude q at a distance r apart in air of permittivity \u025b<sub>\u25e6<\/sub> is\n<\/div>\n<p>A.\u00a0\u00a0\u00a0\u00a0q\u00a0\u00a0\u00a0\u00a0B.\u00a0\u00a0\u00a0\u00a04\u03c0 q\u00a0\u00a0\u00a0\u00a0C.\u00a0\u00a0\u00a0\u00a0r\u025b<sub>\u25e6<\/sub>\u00a0\u00a0\u00a0\u00a0D.\u00a0\u00a0\u00a0\u00a0q<sup>2<\/sup><br \/>\n\t\t\t\t4\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2<\/sup> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u025b<sub>\u25e6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sub>q\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2<\/sup>\n\t\t\t\t<\/li>\n<\/ol>\n<ol>\n<li>\n<div>The net capacitance in the circuit below is(a) 8.0\u03bcF (b) 6.0\u03bcF (c) 4.0\u03bcF (d)2.0\u03bcF\n<\/div>\n<\/li>\n<\/ol>\n<p><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi13.png\" alt=\"\"\/>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u03bcF<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0P\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04\u03bcF<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02\u03bcF<br \/>\n3. \u00a0\u00a0\u00a0\u00a0The magnitude of the electric field intensity at a distance r from a point charge q is\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\nA.\u00a0\u00a0\u00a0\u00a0q<sup>2<\/sup><sup>\u00a0\u00a0\u00a0\u00a0<\/sup>B.\u00a0\u00a0\u00a0\u00a0q\u00a0\u00a0\u00a0\u00a0C.\u00a0\u00a0\u00a0\u00a0q\u00a0\u00a0\u00a0\u00a0D.\u00a0\u00a0\u00a0\u00a0q<br \/>\n\t\t4\u03c0\u025b<sub>\u25e6<\/sub>r\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04\u03c0\u025b<sub>\u25e6<\/sub>r<sup>2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup>4\u03c0\u025b<sub>\u25e6<\/sub>r\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04\u03c0\u025b<sub>\u25e6<\/sub><sup>2<\/sup>r<br \/>\n4.\u00a0\u00a0\u00a0\u00a0Calculate the force acting on an electron carrying a charge of 1.6 x 10<sup>-19<\/sup>C in an electric field of intensity 5.0 x 10<sup>8<\/sup> N\/C is (a) 3.2&#215;10<sup>-29<\/sup>N\u00a0\u00a0\u00a0\u00a0(b) 8.0&#215;10<sup>-11<\/sup> (c) 3.1x 10<sup>27<\/sup>N (d) 4.6&#215;10<sup>-6<\/sup>N<br \/>\n5. Find the electric field intensity in a vacuum at a distance of 10cm from a point charge of 15\u00b5c If 1\/4\u03c0\u025b<sub>\u25e6<\/sub> = 9.0 x 10<sup>9<\/sup><br \/>\n\t\tA.\u00a0\u00a0\u00a0\u00a01.35 x 10<sup>7<\/sup>NC<sup>-1<\/sup>   B. 1.4 x 10<sup>10<\/sup>NC<sup>-1<\/sup>   C. 1.3 x 10<sup>11<\/sup>NC<sup>-1<\/sup> D.1.5&#215;10<sup>10<\/sup>NC<sup>-1<\/sup><br \/>\n\t\tUse the diagram shown below to answer questions 6 and 7<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi14.png\" alt=\"\"\/>6\u03bcF\u00a0\u00a0\u00a0\u00a0 6\u03bcF\u00a0\u00a0\u00a0\u00a0    6\u03bcF<\/p>\n<p>\u00a0<br \/>\n\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0100V<\/p>\n<ol>\n<li>\n<div>What is the effective capacitance in the circuit? (a) 2 \u03bcF (b) 6 \u03bcF (c) 18 \u03bcF (d) 216 \u03bcF\n<\/div>\n<\/li>\n<li>\n<div>What is the total energy store by the capacitors? (a) 2.0&#215;10<sup>-4<\/sup>J (b) 1.0&#215;10<sup>-4<\/sup>J (c) 9.0&#215;10<sup>-2<\/sup>J (d) 1.0&#215;10<sup>-2<\/sup>J\n<\/div>\n<\/li>\n<li>\n<div>Which of the following statements is I are true about an isolated positively charged sphere? (I.) It contains excess positive charges (II.) It has an electric field associated with it. (III.) It carries electric current. (IV.) It has excess negative charges. (a) I and II only (b) I, II and III only (c) II and IV only\u00a0\u00a0\u00a0\u00a0 (d) III and IV (e) I and III only\n<\/div>\n<\/li>\n<li>\n<div>The potential difference across a parallel plate capacitor is 500V while the charge on either plate is 12\u03bcC. Calculate the capacitance of the capacitor (a) 6.0&#215;10<sup>-3<\/sup>F (b) 2.4&#215;10<sup>-4<\/sup>F (c) 6.0&#215;10<sup>-5<\/sup>F (d) 2.4&#215;10<sup>-8<\/sup>F\n<\/div>\n<\/li>\n<li>\n<div>As the plates of a charged variable capacitor are moved closer together, the potential difference between them (a) increases (b) decreases (c) remains the same (d) is doubled\n<\/div>\n<\/li>\n<\/ol>\n<p><strong>SECTION B<\/strong><\/p>\n<ol>\n<li>\n<div>If three charges are distributed as shown in the diagram below.\n<\/div>\n<p><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi15.png\" alt=\"\"\/>     +10C\u00a0\u00a0\u00a0\u00a0         3m\u00a0\u00a0\u00a0\u00a00 \u2013 20C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<br \/>\n2m\n<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+16C<br \/>\nFind the resultant force on the +10C charge (take   1   = 9.0 x 10<sup>9<\/sup> S.I. units)<br \/>\n4\u03c0\u025b<sub>\u25e6<\/sub><\/p>\n<ol>\n<li>\n<div>A capacitor stores 8 x 10<sup>-4<\/sup>c of charge when the potential difference between the plates is 100v. What is capacitance?\n<\/div>\n<p>\u00a0<\/li>\n<li>\n<div>Two charges of +5uc and -10NC are separated by a distance of 8cm in a vacuum as shown below.\n<\/div>\n<p>15uc\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0B\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-10uc\n<\/li>\n<\/ol>\n<p><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi16.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi17.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi18.png\" alt=\"\"\/><br \/>\n\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi19.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi20.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi21.png\" alt=\"\"\/><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0624_Week3SS3Fi22.png\" alt=\"\"\/>             3cm\u00a0\u00a0\u00a0\u00a0            5cm<br \/>\nCalculate the magnitude and direction of the resultant electric field intensity due to point B.<br \/>\n 1 = 9.0 x 10<sup>9 <\/sup>S.I unit<br \/>\n4\u03c0\u025b<sub>\u25e6<\/sub><strong><br \/>\n\t\t\t<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>WEEK 3 Electric Field CONTENTS Electric Field Coulomb&#8217;s Law Electric Field Intensity Electric Potential ELECTRIC&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,303],"tags":[],"class_list":["post-3896","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss3-physics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3896","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3896"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3896\/revisions"}],"predecessor-version":[{"id":3897,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3896\/revisions\/3897"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3896"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3896"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3896"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}