An insulating rod is bent into the semicircle of radius R shown in the figure. T

Question

An insulating rod is bent into the semicircle of radius R shown in the figure. The rod is divided into three sections of equal length. A charge -q is placed on the middle section of the rod, and a charge +2q is placed on both the top and bottom sections of the rod.

Find an expression for the electric field E at the center of the semicircle. Give your answer in component form.

Explain (in words) why your answer to part (a) makes sense based on the arrangement of charge on the semicircle.

Evaluate the field strength if R= 10 cm and q= 10 nC

An insulating rod is bent into the semicircle of radius R shown in the figure. The rod is divided into three sections of equal length. A charge -q is placed on the middle section of the rod, and a charge +2q is placed on both the top and bottom sections of the rod. Find an expression for the electric field E at the center of the semicircle. Give your answer in component form. Explain (in words) why your answer to part (a) makes sense based on the arrangement of charge on the semicircle. Evaluate the field strength if R= 10 cm and q= 10 nC

Explanation / Answer

Draw a horizontal line through O to intersect the rod at P.

Now consider a small part of the rod such that the perpendicular from O to that part is inclined at an angle ? with OP. Let this part subtend an angle d? at the center.

So length of this part = R d?

Charge of this part = R d? * Q / l
where Q and l are the charge & length of the entire rod.

Electric field due to this charge
dE = k*(R d? * Q / l) / R^2 = k Q d? / Rl

The direction of this electric field is along the radius joining that part to the center O.

Now consider another part below OP such that the radius through this part is also inclined to OP at an angle ?.

The electric field exerted by this part will also be dE.

Now the vertical components of fields due to these two mirror elements cancel each other and the net field is along OP.

Similarly, the vertical field due to any part will be canceled by that due to its mirror element.

So net field at O will be the sum of components of fields due to all parts along OPComponent of field due to a part along OP = dE cos? = k Q cos? d? / Rl

For E(net) integrate from ? = -?/2 to ? = ?/2.

E(net) = 2 k Q / R l

Also l = ?R

So
E(net) = 2 ? k Q / l^2

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