EDWARD,SAMU answered A too and left me the B,so i can try it,then if i can't he will came back to it.?

Question

EDWARD,SAMU answered A too and left me the B,so i can try it,then if i can't he will came back to it.?

Answer 1

Are you talking about the following problem?

B) a disk of radius R has a uniform charge per unit area () .calculate the electric field along the axis of the disk.

You have written () as surface charge density. () looks like parentheses. Therefore, I am assuming σ as surface charge density. I do not know how to put integral symbol here. So, I have put the complete sentence integral of from this to that in so many places. Please put the symbol instead of this.

Also draw diagrams to explain the solution.

I suggest that you should learn how to use Calculus to solve the problems in Physics. Then it will be easy for you to sove such problems.

I am giving the solution here. If you want, you can try first yourself and then compare with the following solution.

Solution: -

By symmetry, electric field at the centre of the disk = 0

To find the electric field at any other point on the axis of the disk, let us first find the electric field due to a ring. Suppose that there is a thin ring of radius r. The linear charge density on the ring is q. We have to find the electric field at a point P on the axis of the ring.

Let a = P's distance from the centre of the ring.

Consider an infinitesimal element of length dl on the surface of the ring.

Charge on the element = q*dl

If b = distance of this element fromP, then

b = (r^2 + a^2)^1/2

Electric field at P by this element = dE = (1/4πε0) * q*dl/b^2

dE = (1/4πε0) * q*dl/(r^2 + a^2)

dE = (q/4πε0) * dl/(r^2 + a^2)

The above is the magnitude of electric field. If we take electric fields from all elements and add them, then the result will be along the axis and other components will cancel out. So, we should take component of electric field along the axis of the ring.

That component = dE * a/b = (q/4πε0) * dl/(r^2 + a^2) * a/b

=  (q/4πε0) * dl/(r^2 + a^2) * a/ (r^2 + a^2)^1/2

= (aq/4πε0) * dl/(r^2 + a^2)^3/2

Electric field at P due to the whole ring = integral of  (aq/4πε0) * dl/(r^2 + a^2)^3/2 over the ring

= (aq/4πε0) * 1/(r^2 + a^2)^3/2 * integral of dl over the ring because {qa/(4*π*ε0)}*{1/(r^2 + a^2)^3/2} is a constant.

=  (aq/4πε0) * 1/(r^2 + a^2)^3/2 * circumference of the ring.

=  (aq/4πε0) * 1/(r^2 + a^2)^3/2  *2*π*r

= (aqr/2ε0) * 1/(r^2 + a^2)^3/2     --------------------------------(1)

New consider the disk. Take a ring of inner radius x and infinitesimal width dx, coaxial with the disk. Area of the ring = 2*π*x*dx

Charge on the ring = area of ring * surface charge density =  2*π*x*dx*σ

Length of the ring = 2*π*x

Linear charge density on the ring = charge on the ring/length of the ring = 2*π*x*dx*σ/(2*π*x) = dx*σ

Use this linear charge density in place of q and radius x in place of r in equation (1)

Then the electric field on P due to the ring =  (a*dx*σ*x/2ε0) * 1/(x^2 + a^2)^3/2  

= (a*σ/2ε0) * x*dx/(x^2 + a^2)^3/2

Electric field due to whole disk E = integral of above from 0 to R

Or, E = integral of (a*σ/2ε0) * x*dx/(x^2 + a^2)^3/2 from 0 to R

Let x^2 + a^2 = z^2

Then 2xdx = 2zdz

Or, xdx = zdz

Therefore, E = integral of (a*σ/2ε0) * zdz/z^3 from x = 0 to x = R

Or, E = integral of (a*σ/2ε0) * dz/z^2 from x = 0 to x = R

Or, E = (a*σ/2ε0) * (-1/z) from x = 0 to x = R

Or, E = (a*σ/2ε0) * {-1/(x^2 + a^2)^1/2} from x = 0 to x = R

Or, E = [(a*σ/2ε0) * {-1/(R^2 + a^2)^1/2}] - [(a*σ/2ε0) * {-1/(0^2 + a^2)^1/2}]

Or, E = [(a*σ/2ε0) * {-1/(R^2 + a^2)^1/2}] - [(a*σ/2ε0) * (-1/a)

Or, E = [(a*σ/2ε0) * {-1/(R^2 + a^2)^1/2}] + (σ/2ε0)

Or, E = (σ/2ε0) * {1 - a/(R^2 + a^2)^1/2}

Ans: Electric field at the centre of the disk = 0

Electric field at a distance some other distance a from the centre of the disk

= (σ/2ε0) * {1 - a/(R^2 + a^2)^1/2}

R = radius of the disk.

The field is along the axis.