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1. The electrostatic potential difference between two point embedded in an electric field is b AV = Vab=- JE di a This i

Posted: Mon Jul 11, 2022 1:03 pm
by answerhappygod
1 The Electrostatic Potential Difference Between Two Point Embedded In An Electric Field Is B Av Vab Je Di A This I 1
1 The Electrostatic Potential Difference Between Two Point Embedded In An Electric Field Is B Av Vab Je Di A This I 1 (242.24 KiB) Viewed 33 times
1. The electrostatic potential difference between two point embedded in an electric field is b AV = Vab=- JE di a This is the negative of the work done by the field in moving the charge from point a to point b in the field. It is measured in units of volts. 1 [V] =1[J/C] As long as the field is static, it does not matter what path is taken to evaluate AV. V = 2, The electrostatic potential of a single point charge, as measures at the location away from the charge is just 4TTET. This is a scalar, not a vector! To find the total potential at one must add up the contributions from all charges, which can be expressed as either a sum, if the charges are discrete, or as an integral if the charges are smoothly spread out: qi ATED 7² Fi V (F): charge. pd³ xs V (T) = 4TES -7₁ where dq=charge density x volume element = pd³xs. The subscript s, refers to the distribution of the source charges giving rise to the potential. where is where the potential is measure, and is the location of the ith 3. In either case, the electric field at any point is ovaj ark dy Ē = -32 - In other words the field reflects how quickly the potential changes in space. The set of derivatives taken is called the 'gradient of V', or in conventional notation E = -VV. Think of the gravity analogy. The negative 'Gradient' of the earth's surface gives you the direction that water would flow if you let it go. It is the 'steepest way downhill'. Likewise the E-field points the direction in which the potential changes most quickly.