Tiêu đề 02. ELECTROSTATIC POTENTIAL AND CAPACITANCE.pdf ✅

Electric Field is Conservative In an electric field work done by the electric field in moving a unit positive charge fromone point to the other, depends only on the position of those two points and does notdepend on the path joining them. Electrostatic Potential Electrostatic potential is defined as “Work required to be done against the force by electric field in bringing a unit positive charge from infinite distance to the given point inthe electric field us called the electrostatic potential (V) at that point”. According to above definition the electric potential at point P is given by the formula Vp = −∫ E⃗ p ∞ . ⃗dr⃗⃗⃗ Electric potential is scalar quantity. SI units (J/C) called as volt (V). Potential at a point Due to a Point Charge Consider a point charge positive Q at the origin. To deter let P be the point at a distance ‘r’ from origin of coordinate axis. Since work done in electric field is independent of path, We will consider radial path as shown in figure. According to definition of electric potential we can use the equation Vp = −∫ E⃗ p ∞ . ⃗dr⃗⃗⃗ And electric field E is E⃗ = 1 4πε0 Q r 3 r , given by. Vp = −∫ 1 4πε0 Q r 3 r p ∞ . ⃗dr⃗⃗⃗ Vp = −∫ 1 4πε0 Q r 2 p ∞ . dr Vp = − Q 4πε0 [ −1 r ] Vp = Q 4πε0r Electric Potential Due to Group of Point Charge The potential at any point due to group of point charges is the algebraic sum of thepotentials contributed at the same point by all the individual point charges V = V1 +V2 + V3 + .... Electric Potential Difference Electric potential difference is defined as “Work required to be done to take a unit positivecharge from one point (say P) to another point(say Q) against the electric field According to formula for potential at point P VP = − ∫ p ∞ E⃗ ⋅ ⃗dr⃗⃗⃗ Thus, potential at point Q is given by VQ = − ∫ Q ∞ E⃗ ⋅ ⃗dr⃗⃗⃗ CHAPTER – 2 ELECTROSTATIC POTENTIAL AND CAPACITANCE ELECTR O STA T IC P O T E NTIAL AND CA P ACITA NCE
From above formula potential difference between points Q and P is given by VQ − VP = (− ∫ Q ∞ E⃗ ⋅ ⃗dr⃗⃗⃗ ) −(− ∫ P ∞ E⃗ ⋅ ⃗dr⃗⃗⃗ ) VQ − VP = ∫ ∞ Q E⃗ ⋅ ⃗dr⃗⃗⃗ + ∫ P ∞ E⃗ ⋅ dr⃗⃗⃗⃗ VQ − VP = ∫ P Q E⃗ ⋅ ⃗dr⃗⃗⃗ VQ − VP = − ∫ Q P E⃗ ⋅ ⃗dr⃗⃗⃗ Or work done in moving charge from point P to point Q SI unit of potential is V and dimensional formula [M1 L 2 T −3 A −1 ] Electrostatic Potential Energy The electric potential energy is defined as "The work required to be done against theelectric field in bringing a given charge (q), from infinite distance to the given point in the electric field motion without acceleration is called the electric potential energy of that charge at that point." From definition of electric potential energy and the electric potential we can write electricpotential energy of charge q at point P, as Up = −∫∞ p qE⃗ ⋅ ⃗dr⃗⃗⃗ = q∫∞ p E⃗ ⋅ ⃗dr⃗⃗⃗ = qVp The absolute value of the electric potential energy is not at all important, only the difference in its value is important. Here, in moving a charge q, from point P to Q, withoutacceleration, the work required to be done by the external force, shows the difference in the electric potential energies (UQ - Up) of this charge q, at those two points. UQ − UP = −q∫P Q E⃗ ⋅ ⃗dr⃗⃗⃗ Electric potential energy is of the entire system of the sources producing the field and the Charge, for some configuration, and when the configuration changes the electric potential energy of the system also changes. Potential Energy of a System of Two Point Charges The potential energy possessed by a system of two - point charges q1 And q2 Separated by a distance ' r ' is the work done required to bring them to these arrangements from infinity. This electrostatic potential energy is given by U = q1q2 4πε0r Electric Potential Energy of a System of Point Charges The electric potential energy of such a system is the work done in assembling this system starting from infinite separation between any two-point charges. For a system of point charges q1 , q2 , q3 ... . qn The potential energy is U = 1 2 ∑ n i=1 ∑ n j=1 qiqj 4πε0rij (i ≠ j) It simply means that we have to consider all the pairs that are possible Important points related electrostatic potential energy (i) Work done required by an external agency to move a charge q from A to B in an electric field with constant speed (ii) When a charge q is let free in an electric field, it loses potential energy and gains kinetic energy, if it goes from A to B, then loss in potential energy = gain in kinetic energy q(VB − VA ) = 1 2 mVB 2 − 1 2 mVA 2 Q. Find work done by some external force in moving a charge q = 2μC from infinity to a point where electric potential is 104 V Sol. Work W = Vq = (104 V)(2 × 10−6C) = 2 × 10−2 J Q. The electric field at distance r perpendicularly from the length of an infinitely long wire is E(r) = λ 2πε0r , where λ is the linear charge density of the wire. Find the potential at a point having distance b from the wire with respect to a point having distance a from the wire (a > b) Sol. Let Va be reference point thus Va = 0 Vb −Va = −∫ b a E⃗ ⋅ ⃗dr⃗⃗⃗ Vb −Va = −∫ b a λ 2πε0r dr ∴ (E⃗ ∥ ⃗dr⃗⃗⃗ ) Vb −Va = − λ 2πε0 ∫ b a dr r Vb −Va = − λ 2πε0 [ln r]a b Vb −Va = − λ 2πε0 ln ( a b ) ∵ Va = 0 reference Vba = − λ 2πε0 ln ( a b )

(b) Electric potential at an internal point is given by equation V = 1 4πε0 Q 2R3 [3R 2 − r 2 ] Here R is the radius of the sphere and r is the distance of point from the centre. Relation Between the Electric Field and Electric Potential We know that electric potential from electric field is given by VP = −∫ P ∞ E⃗ ⋅ ⃗dr⃗⃗⃗ And potential difference between two points is given by VQ − VP = −∫ Q P E⃗ ⋅ ⃗dr⃗⃗⃗ If points P and Q are very close to each other, then for such a small displacement ⃗dr⃗⃗⃗ , integration is not required and only term E⃗ ⋅ ⃗dr⃗⃗⃗ can be kept thus dV = −E⃗ ⋅ ̅dr̅̅ i) If ⃗dr⃗⃗⃗ , is the direction of electric field E⃗ , E⃗ ⋅ ⃗dr⃗⃗⃗ = Edrcos θ = Edr dV = −Edr E = − dV dr This equation gives the magnitude of electric field in the direction of displacement ⃗dr⃗⃗⃗ . Here dV dr = potential difference per unit distance. It is called the potential gradient. Its unit is Vm−1 , which is equivalent to N/C. (ii) If ⃗dr⃗⃗⃗ , is not in the direction of E⃗ , but in some other direction, the − dV dr would give us the component of electric field in the direction of that displacement If electric field is in X direction and displacement is in any direction (in three dimensions) then E⃗ = Exıˆ and ⃗dr⃗⃗⃗ = dxıˆ + dyjˆ+ dzkˆ ∴ dV = −(Exıˆ) ⋅ (dxıˆ+ dyjˆ + dzkˆ ) = −Exdx ∴ Ex = − dV dx Similarly, if the electric field is Y and only in Z direction respectively, we would get Ey = − dV dy and Ez = − dV dz Now if the electric field also have three (x, y, z) components, then Ex = − ∂V ∂x . Ey = − ∂V ∂y . Ez = − ∂V ∂z . And E⃗ = −( ∂V ∂x ıˆ+ ∂V ∂y jˆ + ∂v ∂z kˆ ). Here ∂V ∂x , ∂V ∂y , ∂V ∂z shows the partial differentiation of V(x, y, z) with respect to x, y, z respectively. Moreover, the potential differentiation of V(x, y, z) with respect to x means the differentiation of V with respect to x only, by taking y and z in the formula of V as constant Dielectrics and Polarization Dielectrics, in general, can be described as materials that are very poor conductors of electric current. They are basically insulators and contain no free electron. Dielectrics can be easily polarized when an electric field is applied to it. Thus, their behavior in an electric field is entirely different from that of conductors as would be clear from the following discussion. Q. The electric potential in a region is represented as V = 2x + 3y − z. Obtain expression for the electric field strength Sol. We know E⃗ = −( ∂V ∂x ıˆ+ ∂V ∂y jˆ + ∂v ∂z kˆ ) Here ∂V ∂x = ∂ ∂x [2x + 3y − z] = 2 ∂V ∂y = ∂ ∂x [2x + 3y− z] = 3 ∂V ∂x = ∂ ∂x [2x + 3y − z] = −1 E⃗ = 2ıˆ+ 3jˆ− kˆ Q. The electrical potential due to a point charge is given by V = 1 4πs0 Q r . Determine- a) the radial component of the electric field b) the x-component of the electric field Sol. a) The radial component of electric field Er = − dV dr = 1 4πε0 Q r 2 . (b) In terms of rectangular components, the radial distance r = (x 2 + y 2 + z 2 ) 1/2 ; therefore, the potential function V = 1 4πε0 Q (x 2+y 2+z 2) 1 2 To find the x-component of the electric field, we treat y and z constants. Thus Ex = − ∂V ∂x Ex = 1 4πε0 Qx (x 2+y 2+z 2) 3 2 Ex = 1 4πε0 r 3 r 3