In the Figure a nonconducting rod of length L= 8.15 cm has charge –q = -4.23 fC
ID: 2043040 • Letter: I
Question
In the Figure a nonconducting rod of length L= 8.15 cm has charge –q = -4.23 fC uniformlydistributed along its length. (a) What is thelinear charge density of the rod? What are the(b) magnitude and (c) direction(relative to thepositive direction of the s-axis) of the electricfield produced at point P, at a distance a = 12.0cm from the rod? What is the electric fieldproduced at point P, at distance a = 12.0 cm fromthe rod? What is the electric field magnitudeproduced at distance a = 50 m by (d) the rod and(e) a particle of charge –q = -4.23 fC thatreplaces the rod?7)
Explanation / Answer
In the figure below, a nonconducting rod of length L = 8.00 cm has charge q = -4.68 fC uniformly distributed along its length. (a) What is the linear charge density of the rod? (b) What is the magnitude of the electric field at point P, a distance a = 12.0 cm from the end of the rod? (c) What is its direction? (d) What is the electric field magnitude produced at distance a = 50 m by the rod? (e) Repeat part (d) for a particle of charge -q = -4.68 fC that replaces the rod. solution Electric Field due to a Charge Distribution The electric field at a point due to a distribution of charge of any shape can be calculated through the integration of what is known as charge density. There are three different possibilities for charge density, dependent on whether the charge is spread over a line, a flat surface, or a volume. The common symbols used for charge density are the following: Linear Charge Density ?=q/L Surface Charge Density s=q/A Volume Charge Density ?=q/V The electric field is found by integrating the charge density and the separation r of the point from the charged object. The integral takes the form: E=?k/r^2 dq where k=8.99E9 r is the separation of a point on the charged object and the point where the electric field is to be measured. Note that r is often different for different points on the charged object, and this variable r is what leads to the electric field requiring an integral calculation. dq is the small amount of charge at a given point on the object. It will be one of the three possibilities dependent on the type of charge distribution. Linear Charge Density dq=? ds Surface Charge Density dq=s dA=s L ds Volume Charge Density dq=? dV Now to solve the problem with numbers: We are given the following information L=8.00 cm=8.00E-2 m=0.08 m q=-4.68 fC=-4.68E-15 C The first step is to calculate the linear charge density ?=q/L ?=-4.68E-15/8.00E-2=-5.85E-14 C/m (a) And then the electric field at point P through integration E=?k/r^2 dq=?k/r^2 ? ds=k??1/r^2 ds where r is the separation of a point on the charged rod and the point P. Notice this r will take on all values between r=L+a for the extreme left end of the rod r=a for the extreme right end of the rod At this point you have the freedom of choosing where to set the origin in the picture, though there are some locations that are better than the others for simplifying the computations involved. The end of a line of charge where point P is located is often the best choice for the origin so that we can keep the relationship between r and x as simple as possible. The relationship between ds and x is simply ds=dx. The most challenging part in the set up of integrals of this type is determining the relation between r and x. If the distance x is measured from the right end of the rod, the distance between a point on the rod at a location x from the right end and the point P will be r=x+a. These results can now be put into the integral E=?k/r^2 dq=?k/r^2 ? ds=k??1/r^2 ds=k??1/(x+a)^2 dx An integral of this type can be computed using u substitution where u=x+a du=dx E=k??1/(x+a)^2 dx=k??1/(u)^2 du=k??u^-2 du=k? u^-1/-1 dx=k? -1/u=-k?/u=-k?/(x+a) All that is left is to plug in the bounds and then evaluate with the given numbers. The bound for this rod will be from the right end x=0 since that is where we set the origin and the left end x=L. E=-k?/(x+a) x evaluated from x=0 and x=L E=[-k?/(L+a)]-[-k?/(0+a)]=-k?/(L+a)+k?/(a)=k?[-1/(L+a)+1/(a)]=k?[1/(a)-1/(L+a)] When a=12.0 cm=12.0E-2 m=0.12 m E=k?[1/(a)-1/(L+a)]=8.99E9*5.85E-14*[1/(0.12)-1/(0.08+0.12)]=8.99E9*5.85E-14*[1/(0.12)-1/(0.20)]=1.75305E-3 N/C (b) Always use the absolute value of the charge density ? for the magnitude of the electric field. The direction can be found from the sign of the charged object. Since it is negative, the electric field will point towards it. At point P, this direction will be to the left. Another way of stating to the left is 180 deg from the positive x axis (which itself is to the right). (c) When a=50 m E=k?[1/(a)-1/(L+a)]=8.99E9*5.85E-14*[1/(50)-1/(0.08+50)]=8.99E9*5.85E-14*[1/(50)-1/(50.08)]=1.680240E-8 N/C (d) If instead the rod were a point charge at a distance a=50 m we can use the point charge electric field formula, which does not require an integral. E=k q/r^2=(8.99E9*4.68E-15)/(50)^2=1.682928E-8 N/C (e) Notice that the formula for the rod at a large distance a in answer (d) compared to L produces about the same electric field value as the point charge formula in answer (e). This is true for all charged objects. The further away from the charged object that you measure the electric field, the closer the value will be equivalent to the charged object being a point charge. The reason for this is simple: The further away from an object you are, the more it appears as a small point.