University:
Kentucky State University
Course:
PHY 212 | General Physics II
Academic year:

2022
Views:

48

Pages:

7
Author:

Tyshawn Shaffer

The Electric Field • a charged particle Q creates a “field” in all space: E + • other charged particle (q) does sense the field and does experience a force in response. Definition of electric field: F = qE F & E are parallel for +ve q F & E are antiparallel for –ve q This is the force F exerted by the field E (created by Q) on the charge q. It is a vector equation that tells you magnitude and direction of the force. The units of electric field are newtons/coulomb. F0  N  E  = = q0  C The direction of the electric field is radially outward (away) from positive charge and radially inward (towards) negative charge. + The Electric Field Due to a Point Charge Coulomb's law says q1q 2 F =k 2 , 12 r12 If q1 creates an electric field, and q2 enters this field, then the electric field due to the point charge q1 (the source) is the electric force F12 divided by q2 q1 |E q1 |=k 2 r12 or, generally q E=k 2 r If we define r̂ as a unit vector from the source point to the field point… source point r̂ + field point …then the vector equation for the electric field of a point charge becomes: q E=k 2 rˆ r r̂ Radially outward for +ve q Radially inward for –ve q Example: calculate the magnitude of the electric field at the electron’s distance away from the proton in a hydrogen atom (5.3x10-11 m). +e + -e D EP 9 19 k q k(+e) 9 10 1.6 10  N EP  2   2 11 2 r D 5.3 10  C E P  5.110 11 N C Quiz: What is the force exerted by this field on the electron? Example: calculate the electric field at point P, which lies on the perpendicular bisector a distance L from the line d. y E+   P r +q + E = E  E L  E y = 0 (by symmetry) E x = 2E,x (by symmetry) Er E x = +2E  cos   d - -q x Example: calculate the electric field at point P, which lies on the perpendicular bisector a distance L from the line d. y   P r +q + E x = +2E  cos  E+ L  d/2 d/2 d E x = +2E  = +E  r r Er d/2  -q d Ex x k q d kqd =+ 2 = 3 r r r qd ˆ E = i 3 40 r