Electrical Engineering ⇒ Topic : Electric Field Intensity due to a Point Charge at Rest

April
 
Electric Field Intensity due to a Point Charge at Rest Consider a charge q placed in free space. Let Δq be the test charge placed at a distance d from q. The elementary force acting on Δq due to q is ΔF, which is given as ΔF = q Δq/4πε_{0} r^{2 }acting along the line joining q and Δq. Therefore .............. (1) i.e. ................. (2) The electric field intensity at a point where q is placed is defined as the force per unit charge and therefore ..................... (3) Electric field intensity (E) is a vector since F is a vector, and is expressed as Newton/Coulomb (N/C). This suggests that if E is known at any point in the medium, the electrostatic force can be computed, since F =E x q .............(4) Therefore, electric field intensity due to a point charge q which is at a distance r from the charge is given as .............. (5) and acts in a direction away from q. If the charge q is negative, the direction of E is reversed and it acts in the direction of the charge  
 
!! OOPS Login [Click here] is required for more results / answer