85. Electric Field and Electric Potential

85. Electric Field and Electric Potential - ⇔ V x =-Z x x...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Electric Field and Electric Potential Determine the field or the potential from the source (charge distribution): ~ E = 1 4 π² 0 Z dq r 2 ˆ r r dE dV r dq V = 1 4 π² 0 Z dq r Determine the field from the potential: ~ E = - ∂V ∂x ˆ i - ∂V ∂y ˆ j - ∂V ∂z ˆ k Determine the potential from the field: V = - Z ~ r ~ r 0 ~ E · d~ s Systems with uniaxial symmetry: E x ( x ) = - dV dx
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ⇔ V ( x ) =-Z x x E x dx • Application to charged ring: E x = kQx ( x 2 + a 2 ) 3 / 2 ⇔ V = kQ √ x 2 + a 2 • Application to charged disk (at x > ): E x = 2 πσk » 1-x √ x 2 + R 2 – ⇔ V = 2 πσk h p x 2 + R 2-x i tsl85 – p.1/1...
View Full Document

This note was uploaded on 04/30/2008 for the course PHYS 204 taught by Professor Andrevantonder during the Spring '07 term at Rhode Island.

Ask a homework question - tutors are online