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Hi, I have been struggling with this problem. I have no idea what to do. Any help would be greatly appreciated. A visual representation of the problem is provided.

Consider a semi-circular arc of radius R = 0.20 cm and linear charge density lamda(theta) = 10 cos(theta) expressed in C/m. Note that the linear charge density on an arc is defined as the charge per unit arc length.

(a) Consider a small element of arc angle delta theta and position theta. What is the charge delta q of this element? (Recall that arc length delta s is given by delta s = R(delta theta) Express your answer in terms of theta and delta theta.

(b) What is the total charge on the arc? Express your answer numerically.

(c) What are (i) the electric field delta E, (ii) the electric potential delta V, at P due to delta q? Express your answer in terms of theta and delta theta.

(d) What are (i) the net electric field E, (ii) the net electric potential V, at P due to the entire arc? Use symmetry wherever possible, and evaluate your final integrals. Express your answer numerically

Hello Tommy,

a. So the arc length is R*DeltaTheta. The cosine function does not plot as a straight line, but since this is a small element of the arc, it can be approximated as linear. It appears from the figure that Theta is in the center of the arc length DeltaTheta. Therefore the charge density at the center of the element, 10 cos(Theta), can be approximated as equal to the average density over the length of the element. If you multiply a linear charge density times the length of the linear distribution, the result has the unit Coulomb. Since the charge density at the center of the element is 10*cos(Theta), you can consider the charge DeltaQ to be

10*cos(Theta)*0.002*DeltaTheta = 0.2*cos(Theta)*0.002*DeltaTheta.

b. The cosine function varies from 1 to 0 as theta increases from 0 to 90 and continues from 0 to -1 as theta increases to 180. Any element, a, in the upper part of the figure has an equal sized element, z, in the lower part of the figure with negative charge such that element z cancels element a. Therefore the net charge on the arc is zero.

ci. The field E due to the charge DeltaQ obeys the basic formula

E = k*q/r^2

So when q = DeltaQ,

E = k*DeltaQ/R^2

E = 9*10^9 N.m^2/C^2 * 10*cos(Theta)*0.002*DeltaTheta / (0.002 m)^2

E = 2.25*10^16*cos(Theta)*DeltaTheta N/C

cii. The potential V due to the charge deltaQ obeys the basic formula

V = k*q/r

So when q = deltaQ,

V = k*DeltaQ/R

V = 9*10^9 * 10*cos(Theta)*0.002*DeltaTheta / 0.002 m

V = 4.5*10^13 cos(Theta)*DeltaTheta N.m/C =

Note that the N.m/C is equivalent to Volts.

di. Consider 2 small elements, one, a, at Theta, which is less than 90 degrees and the other, z, at 180 - Theta. Theta will be limited to 90 degrees so that 180-Theta will cover the remainder of the arc. This will allow the application of symmetry. For description purposes, let's use directions of the compass to describe the direction a vector is pointing. So a vector from the 90 point of the arc pointing toward P will be pointing east. At P, the field, Ea, due to element a will point some amount south of east. The angle between east and Ea will be the angle theta. At P, the field, Ez, due to element z will point the angle theta south of west. You will need to draw these for understanding.

Now we will consider the components of Ea and Ez that point east-west and north-south. Notice that Ea has a component that points east and that Ez has a component pointing west. They cancel. In the north-south direction, Ea and Ez have equal components both pointing south. Therefore if you integrate between 0 and 90 only, and multiply by 2, you will have the answer. From part ci above,

Ea = 2.25*10^16*cos(Theta)*DeltaTheta N/C

The southward component of that is Ea*cos(Theta)

So the function to be integrated is

E = 2*2.25*10^16*cos^2(Theta)*dTheta

I will leave the math to you. But here's a hint:

http://www.mathtutor.ac.uk/integration/integrationusingtrigonometricformulae/tex

Look at page 2.

dii. The approach is similar here but since potential is not a vector, the need to consider north-south and east-west components is eliminated.

I hope this helps,

Steve

Physics

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