Tiêu đề 02. ELECTROSTATIC POTENTIAL AND CAPACITANCE.pdf ✅

Electric Potential Due to Continuous Charge Distribution The electric potential due to continuous charge distribution is the sum of potential of all the infinitesimal charge elements in which the distribution may be divided V = ∫ Dv = ∫ dq 4πε0r Electric Potential Due to a Charged Ring A charge Q is uniformly distributed over the circumference of a ring. Let us calculate the electric potential at an axial point at a distance r from the centre of the ring. The electric potential at P due to the charge element dq of the ring is given by dV = 1 4πε0 dq Z = 1 4πε0 dq √R2 + r 2 Hence electric potential at P due to the uniformly charged ring is given by V = ∫ 1 4πε0 dq √R2 + r 2 = 1 4πε0 1 √R2 + r 2 ∫ dq V = 1 4πε0 Q √R2 + r 2 Electric Potential Due to a Charged Disc at a Point on The Axis A non-conducting disc of radius ' R ' has a uniform surface charge density σC/M2 To calculate the potential at a point on the axis of the disc at a distance from its centre. Consider a circular element of disc of radius X ′ And thickness dx. All points on this ring are at the same distance Z = √x 2 + r 2 , from the point P. The charge on the ring is dq = σA dq = σ(2πxdx) and so the potential due to the ring is dV = 1 4πε0 dq Z = 1 4πε0 σ(2πxdx) √x 2 +r 2 Since potential is scalar, there are no components. The potential due to the whole disc is given by V = σ 2ε0 ∫ R 0 x √x 2 + r 2 dx = σ 2ε0 [(x 2 +r 2 ) 1/2 ] 0 R V = σ 2ε0 [(R 2 + r 2 ) 1/2 −r] V = σ 2ε0 [r ( R 2 r 2 + 1) 1/2 − r] For large distance R/r ≪ 1 thus r ( R 2 r 2 + 1) 1/2 ≈ r (1+ R 2 2r 2 ) Substituting above value in equation for potential V = σ 2ε0 [r (1+ R 2 2r 2 )− r] V = σ 2ε0 R 2 r 2r 2 V = σ 2ε0 R 2 2r ( π π ) Since Q = oπR 2 V = 1 4πε0 Q r Thus, at large distance, the potential due to disc is the same as that of point charge Electric Potential Due to A Shell A shell of radius R has a charge Q uniformly distributed over its surface. (a) At an external point At point outside a uniform spherical distribution, the electric field is E⃗ = 1 4πε0 Q r 2 rˆ Since E⃗ Is radial, E⃗ ⋅ ⃗dr⃗⃗⃗ = Edr Since V(∞) = 0, we have V(r) = − ∫ r 0 E⃗ ⋅ ⃗dr⃗⃗⃗ = − ∫ r 0 Q 4πε0r 2 dr = − Q 4πε0 [ −1 r ] ∞ V(r) = Q 4πε0r (r > R) Thu potential due to uniformly charged shell is the same as that due to a point charge Q at the Centre of the shell. (b) At an internal point At point inside the shell, E = 0. So, work done in bringing a unit positive charge from a point on the surface to any point inside the shell is zero. Thus, the potential has a fixed value at all points within the spherical shell and is equal to the potential at the surface. Above results hold for a conducting sphere also whose charge lies entirely on the outer surface. Electric Potential Due to A Non-Conducting Charged Sphere A charge Q is uniformly distributed through a non- conducting volume of radius R. (a) Electric potential at external point is given by equation. ' r ' is the distance of point from the center of the sphere V = 1 4πε0 Q r Q. An electric field is represented by E⃗ = Axıˆ, where A = 10 V/m2 . Find the potential of the origin with respect to the point (10,20)m Sol. E⃗ = Axıˆ = 10xıˆ V(0,0) − V(10,0)) = −∫ (0,0) (10,20) E⃗ ⋅ ⃗dr⃗⃗⃗ V(0,0) − V(10,0)) = −∫ (0,0) (10,20) (10xıˆ) ⋅ (dxıˆ+ dyjˆ) V(0,0) − V(10,2)) = −∫ 0 10 (10xdx) V(0,0) − V(10,00) = −10 [ x 2 2 ] 10 V(0,0) − V(10,0)) = [0 −(−500)] V(0,0) − V(10,0) = 500 V Since V (10,20) is to be taken zero V (0, 0) = 500 volts
(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