Question

Charge of uniform density 50 nC/m3 is distributed throughout the inside of a long nonconducting cylindrical rod (radius = 5.0 cm). Determine the magnitude of the potential difference of point A (2.0 cm from the axis of the rod) and point B (4.0 cm from the axis).

Answer 1

Answer:

magnitude of the potential difference is  -1.7 V

Explanation:

given data

Charge of uniform density σ = 50 nC/m³ = 50 × $10^{-19}$ C/m³

radius = 5.0 cm = 0.05 m

point A = 2 cm = 0.02 m

point B = 4 cm = 0.04 m

solution

as we know inside rod, we construct a cylindrical Gaussian surface of length L and radius r  around the axis.

so  electricfield is always perpendicular to the lateral surface area -2πrL

so that

E = $\frac{Q_in}{A\epsilon _o}$   ..........1

E = $\frac{\rho \pi r^2 l}{2\pi r l \epsilon _o}$    

E = $\frac{\rho r}{2\epsilon _o}$  

Δv = - ∫ E.dr

Δv =     $\int\limits^{r2}_{r1} {x} \, \frac{\rho r }{2\epsilon _o } dr$      

put here value

Δv =     $\frac{\rho }{2\epsilon _o } \int\limits^{0.04}_{0.02} {x} \, r dr$    

Δv = $\frac{\rho }{2\epsilon _o } \times [\frac{r^2}{2}]^{0.04}_{0.02}$  

Δv = $\frac{50\times 10^{-9} }{4\epsilon _o } \times {0.04^2}- {0.02^2}$  

Δv  = -1.7 V