1. Consider a finite line charge of uniform charge density l and of length L lying along the x-axis from to x=0. Find the field at a point along the axis at x=d.

Take an element pf length dx at a distance x from the origin. Since the point is at a distance d along the axis, the distance from this element is (x+d). The field is thus given by

2.Find the point(s) on the axis of a charged ring where the electric field has maximum strength.

From the expression derived in Example 2, the field on the axis of a charged ring at a distance x is given by . Differentiate this expression and set the result to zero to obtain the maxima at , the two signs indicate that the field becomes maximum to either side of the ring.

3. Find the electric field on the axis of a 600 arc of a circle of radius R with a uniform charge density . Leave the result in terms of an integral. In particular, find the field at the centre of the circle of which the arc is a part.

In doing this problem it is important to realize that the axis of an arc is in the same plane as the arc (unlike the case of a ring whose axis is perpendicular to the plane of the ring). Consider the field at a distance x from the centre of the circle. If we take an element of length at an angle , the point P is at a distance . The y-component of the field

cancels by symmetry. The x-component is given by

The field at the point O is obtained by taking x=0. It gives

4. Determine the electric field at the centre of the base of a semicircle of radius R which has a charge density , where q is the angle made by an element on the semicircle with the base.

Take an element on the semicircle at an angle q with respect to x axis. The strength of the electric field at O due to the charge element is directed radially outward. We can resolve this along x and y axes.

5. Two semi-infinite charged lines lie in the x-y plane making an angle of 1200 with each other. They are smoothly joined by an arc of a circle of radius R. The linear charge density is uniform throughout. Find the electric field at the centre of the circle of which the arc is a part.

We have seen in the Tutorial problem 4 that the field at a perpendicular distance y from the edge of a semi-infinite line has equal components perpendicular to the line and parallel to it (but in opposite direction), each component being . In the figure below we show the field directions at P due to two semi-infinite wires, the directions shown in red is due to the line below the x-axis and that shown in blue is due to the wire above the x-axis. The angle between A1P and B1P is given as 600 and distance of P from each of the wires is R. Considering one of the

wires (say BN), we resolve the fields along x and y directions :

If we consider the wire AM, we get similar expressions but the y-components are in the reverse to direction to that due to BN while the x components add up. The resulting contribution due to the two lines is

To complete the problem we have to add the contribution due to the 600 arc. We have seen in Problem 3 that the field due the arc is along its axis which in this case is the x axis and has a magnitude given by . Adding all contributions we have .