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Physics Question on electrostatic potential and capacitance

A thin non-conducting rod of length 50 cm50 \,cm has a positive charge of uniform linear density 10−12 C/m10^{-12} \,C / m. Find the electric potential due to the rod at a point, which is at a perpendicular distance of 1.0 cm1.0 \,cm from one-end of the rod (ε0=8.8×10−12F/m)\left(\varepsilon_{0}=8.8 \times 10^{-12} F / m \right).

A

0.02 V

B

0.04 V

C

0.06 V

D

1.02 V

Answer

0.04 V

Explanation

Solution

The given, λ=10−12 C/m\lambda=10^{-12}\, C / m,
l=50 cml=50\, cm,
r=1×10−2 mr=1 \times 10^{-2} \,m
and ε0=8.8×10−12 F/m\varepsilon_{0}=8.8 \times 10^{-12} \,F / m
We know that, λ=Ql\lambda=\frac{Q}{l}
10−12=Q50 cm10^{-12}=\frac{Q}{50\, cm }
or 10−12=2Q2×50 cm10^{-12}=\frac{2 Q}{2 \times 50\,cm }
10−12=2Q100 cm,10^{-12}=\frac{2 Q}{100\, cm },
10−12=2Q1 m10^{-12}=\frac{2 Q}{1 \,m},
Q=12×10−12 CQ=\frac{1}{2} \times 10^{-12} \,C,
Electric potential V=14πε0⋅QrV=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r}
=14×3.14×8.8×10−12⋅12×10−121×10−2=\frac{1}{4 \times 3.14 \times 8.8 \times 10^{-12}} \cdot \frac{\frac{1}{2} \times 10^{-12}}{1 \times 10^{-2}}
=100221.05=0.4 V=\frac{100}{221.05}=0.4 \,V