Answer:
[tex]V(b)-V(a) = - \int\limits^b_a {E} \, dL[/tex]
Explanation:
Solution:
- The Electric potential V is given as line integral(L) of Electric Field E. The expression.
[tex]V = -\int\limits^r_0 {E} \, dL[/tex]
Where, r is the radial distance from the axis of charged body.
- The coaxial cable has inner radius (a) and outer radius (b). The potential difference between the inner radius (a) and outer radius (b) can be expressed as follows:
[tex]V(b)-V(a) = - \int\limits^b_0 {E} \, dL + \int\limits^a_0 {E} \, dL \\\\V(b)-V(a) = - \int\limits^b_0 {E} \, dL - \int\limits^0_a {E} \, dL\\\\V(b)-V(a) = - \int\limits^b_a {E} \, dL[/tex]