PHYAP 03 Continuous Charge Distribution By Squadron Leader

PHY-AP -03 Continuous Charge Distribution By Squadron Leader Zahid Mir CS&IT Department , Superior University

Continuous Charge Distribution If the charge is spread uniformly over the surface or volume of the object then it is said to be continuous charge distribution.

Limitation of Coulomb’s Law can apply only to point charges.

Point Charge A charged objects whose size are much smaller then the distance between them.

How to solve the problem? • We cannot apply Coulomb’s law in its point charge form to calculate force exerted by one charged rod to another. • One approach is to consider the rod is covered with point charges and then calculate the force exerted by each point charge of one rod to each point charge of the other—but its very complicated.

Second Approach !! • We assume that charge is a continuous property. • We then divide the charge into infinitesimal elements and then calculate the force due to all the e. • If the object carries a charge ‘q’ then we divide it into small elements each of charge ‘dq’.

Charge Density • It describes how charge is distributed over the length, area or volume (depending upon the dimensions) of the object. • In uniform charge distribution, the value of charge density remains same everywhere of the object.

Charge Density(cont) Linear Charge Density = dq dx Surface Charge Density = dq d. A Volume Charge Density = dq d. V

Evaluation of Continuous Charge Distribution • Imagine the continuous charge distribution to be divided into large number of small charge elements. • Pick an arbitrary charge element and express its charge ‘dq’ in term of line, area, or volume charge density. • Charge element ‘dq’ can be treated as point charge. • Express the force element ‘d. F’ exerted by the charge ‘dq’ on the charge ‘qo’ in terms of Coulomb’s Law. • By taking into account the signs and locations of charge element ‘dq’ and ‘qo’ we will determine the direction of force element ‘d. F’. • The total force is then found by adding all the infinitesimal force elements.

Application A Uniform Line of Charge Linear Charge Density = d. F = 1 4 ∏Є o qo dq r 2 q L There is no effect of force along X-axis Fx = 0 z-axis + dq + dz + + + L + + x-axis + + + r Ѳ y qo y-axis

A Uniform Line of Charge (cont) z 1 r d. Fz 2 Ѳ O Ѳ Ѳ Ѳ r z 2 d. Fz 1 Y-axis d. F 1 Symmetry Argument • For every charge element ‘dq’ located at position +z 1, there is another charge element ‘dq’ located at –z 2. • When we add the forces due to the charge element at +z 1 & -z 2 we find the z-components have equal magnitudes but opposite directions; So their sum is zero. Fz = 0

A Uniform Line of Charge (cont) d. Fy = d. F CosѲ dq = λ dz z Ѳ r 2 = y 2 + z 2 d. Fy = r y 1 q 0 λ dz 4 ∏Є o (y 2 + z 2) CosѲ = y r y √ y 2 + z 2

L/2 Fy = 1 4 ∏Є o q 0 λy dz (y 2 + z 2)3/2 -L/2 q 0 q y√ y 2 + L 2/4 As y >> L Fy = 1 4 ∏Є o q 0 q y 2