In Fig. 24-31a, what is the potential at point P due to charge Q at distance R from P? Set at infinity. (b) In Fig. 24-31b, the same charge has been spread uniformly over a circular arc of radius R and central angle 40. What is the potential at point P, the center of curvature of the arc? (c) In Fig. 24-31c, the same charge Q has been spread uniformly over a circle of radius R . What is the potential at point P , the center of the circle? (d) Rank the three situations according to the magnitude of the electric field that is set up at P, greatest first.

Figures 24-31a, 24-31b and 24-31c are given to describe the point and charge locations for the three given cases.

The electric potential of the system can be given by the amount of work done through force to bring the charge from one position to the other. The difference in the potential can also be given by the integral change in the path due to a constant electric field present in the system.

Formulae:

The electric potential at a point due to an individual charge,

….. (i)

Here, K is the Coulomb’s constant, Q is the charge, and is the distance.

The electric field at a point,

….. (ii)

The line charge density of a material,

….. (iii)

Here, L is the length.

The electric field at a point can be given using equation (ii) as:

Using equation (ii) in equation (i), the potential at point P due to charge Q at distance R from P when brought from infinity can be given as follows

Hence, the value of the potential is .

Suppose the charge per unit length of the given arc is $\lambda$.

Now, consider a small charge element on the arc and that can be given using equation (iii) as follows:

dq= $\lambda Rd0$ ….. (iv)

Here, the length of the arc is,

L=Rd0

If $\alpha$be the full angle of the arc that is given ${40}^0$in this case, then the lines charge density of the arc using equation (iii) becomes:

$\lambda =Q\backslash R\alpha$ ….. (v)

Here, the length of the arc is,

$L=R\alpha$

Now, using equation (v) in equation (iv), the charge value can be given as:

Now, let us consider two symmetrical small charge elements that are located equally at an angle on the arc. Being symmetrical their vertical component cancels out and thus, only horizontal add up.

Thus, the net electric field can be given using the above charge value and equation (ii) as follows: (as the arc has angle $\alpha$, then the angle value for the arc can be given from to $\alpha \backslash 2$.

to $+\alpha /2)$

Thus, the electric field at angle can be given as:

As known that,40=0.68813 rad

Therefore,

Now, the potential at the point P for the arc can be given using equation (i) , equation (a) and the charge value as follows:

From equation (v) you can substitute Q\R for $\lambda \alpha$in the above equation.

$V_3=K(Q\backslash R)$

Hence, the value of the electric potential is K(Q\R) .

In a given circle, the field produced by a charge element is cancelled by another charge element located diametrically opposite to the former.

Thus, the electric field at the center of the circle is zero. Therefore,

$E_3$=0

The potential of a uniform distribution over a circular disc is spread over the circle is independent of angle subtended.

Hence, the potential at the center of circle is K(Q\R) .

From the above calculations, the rank of the situations according to their electric fields is found to be:

Hence, the required rank is .