Title 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

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