Question 4.11: A point charge of 5nC is located at (-3, 4, 0), while line y...

Let the potential at any point be

V=V_{Q}+V_{L}

where V_{Q} and V_{L} are the contributions to V at that point due to the point charge and the line charge, respectively. For the point charge

V_{Q}=-\int E\cdot dl=-\int \frac{Q}{4\pi\varepsilon_{o}r^{2}}a_{r}\cdot dr a_{r}=\frac{Q}{4\pi\varepsilon_{o}r}+C_{1}

For the infinite line charge

V_{L}=-\int E\cdot dl=-\int \frac{\rho_{L}}{2\pi\varepsilon_{o}\rho}a_{\rho }\cdot d\rho a_{\rho }=-\frac{\rho _{L}}{2\pi\varepsilon_{o}}\ln\rho +C_{2}

Hence

V=-\frac{\rho _{L}}{2\pi\varepsilon_{o}}\ln\rho+\frac{Q}{4\pi\varepsilon_{o}r}+C

where C=C_{1}+C_{2} constant, \rho is the perpendicular distance from the line y=1,  z=1 to the field point, and r is the distance from the point charge to the field point.

(a) If V=0 at O(0,0,0), and V at A(5,0,1) is to be determined, we must first determine the values of \rho and r at O and A. Finding r is easy; we use eq. (2.31).

d_{2}=\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}   (Cartesian)

To find \rho for any point (x,y,z), we utilize the fact that \rho is the perpendicular distance from (x,y,z) to line y=1,  z=1, which is parallel to the x-axis. Hence \rho is the distance between (x,y,z) and (x,1,1) because the distance vector between the two points is perpendicular to a_{x}. Thus

\rho =\left|\left(x,y,z\right)-\left(x,1,1\right) \right|=\sqrt{\left(y-1\right)^{2}+\left(z-1\right)^{2} }

Applying this for r and eq. (2.31) for r at points O and A, we obtain

\rho _{O}=\left|\left(0,0,0\right)-\left(0,1,1\right) \right|=\sqrt{2}

r_{O}=\left|\left(0,0,0\right)-\left(-3,4,0\right) \right|=5

\rho _{A}=\left|\left(5,0,1\right)-\left(5,1,1\right) \right|=1

r_{A}=\left|\left(5,0,1\right)-\left(-3,4,0\right) \right|=9

Hence

V_{O}-V_{A}=-\frac{\rho _{L}}{2\pi\varepsilon_{o}}\ln\frac{\rho _{O}}{\rho _{A}}+ \frac{Q}{4\pi\varepsilon_{o}} \left[ \frac{1}{r_{O}}-\frac{1}{r_{A}} \right]

=\frac{-2\cdot 10^{-9}}{2\pi\cdot\frac{10^{-9}}{36\pi}}\ln\frac{\sqrt{2}}{1}+ \frac{5\cdot 10^{-9}}{4\pi\cdot\frac{10^{-9}}{36\pi}} \left[ \frac{1}{5}-\frac{1}{9} \right]

0-V_{A}=-36\ln\sqrt{2}+45\left(\frac{1}{5}-\frac{1}{9}\right)

or

V_{A}=36\ln\sqrt{2}-4=8.477V

Notice that we have avoided calculating the constant C by subtracting one potential from another and that it does not matter which one is subtracted from which.

(b) If V=100 at B(1,2,1) and V at C(-2,5,3) is to be determined, we find

\rho _{B}=\left|\left(1,2,1\right)-\left(1,1,1\right) \right|=1

r_{B}=\left|\left(1,2,1\right)-\left(-3,4,0\right) \right|=\sqrt{21}

\rho _{C}=\left|\left(-2,5,3\right)-\left(-2,1,1\right) \right|=\sqrt{20}

r_{C}=\left|\left(-2,5,3\right)-\left(-3,4,0\right) \right|=\sqrt{11}

V_{C}-V_{B}=-\frac{\rho _{L}}{2\pi\varepsilon_{o}}\ln\frac{\rho _{C}}{\rho _{B}}+ \frac{Q}{4\pi\varepsilon_{o}} \left[ \frac{1}{r_{C}}-\frac{1}{r_{B}} \right]

V_{C}-100=-36\ln\frac{\sqrt{20}}{1}+ 45\cdot \left[ \frac{1}{\sqrt{11} }-\frac{1}{\sqrt{21} } \right]=-50.175V

or

V_{C}=49.825V

(c) To find the potential difference between two points, we do not need a potential reference if a common reference is assumed.

V_{BC}=V_{C}-V_{B}=49.825-100=-50.175V