In Fig. 22-44, a thin glass rod develops a semicircle that radius r= 3.87 cm. Fee is uniformly dispersed along the rod, v +q =3.52 computer in the upper fifty percent and -q = -3.52 pc in the lower half. Whatis the magnitude of the electrical field at P, the facility of thesemicircle? The main principle used is to settle this difficulty is electrical field in ~ the facility of a one arc.

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First calculate the electric field due to each 4 minutes 1 is found at the center.

Later, take the vector amount of electric field due to two 4 minutes 1 to calculate the net electric field.

Finally combine the electric field expression end the complete arc and also plug in the values to calculation the size of electrical field.

The electrical field at the facility of curvature because of charge circulation on a one arc is offered by following expression: Here, is the permittivity that the cost-free space, is the size of charge, is the distance of the suggest from the charge, and also is the angle.

Net electric field in ~ the center is provided by vector sum of electrical field because of two quarters as follows: Here, and are the electrical field in ~ the center.

Find the electrical field at the facility of circular arc as result of upper half.

Consider a small line aspect of angular width making angle with hopeful y- axis through uniform direct charge density .

Let q be the fee at the center of circular arc of radius r.

The electrical field early out to tiny line element is, Linear charge thickness is provided as follows: So, the electrical field due to little element is, The horizontal component of electrical field point out in positive x-direction and the upright component points in an adverse y- direction.

In vector form, electrical field is offered below: As upper and also lower fifty percent of the arc are symmetric through respect come horizontal, having actually same charge however of the contrary sign, the electrical field because of lower arc is given below: Net electrical field is given by the vector amount of the two areas as follows: Find the electrical field at the center of the semicircle.

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Net electrical field at the center due to little elements is, Total electrical field due to whole arc is, Substitute because that , for q , because that , because that and for r together follows: Remove the suffices that the systems as follows: The electric field action in an adverse y- direction and also the magnitude of the electrical field is .