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Physics Question on Electric charges and fields

A rod of length 2.4 m2.4\, m and radius 4.6 mm4.6\, mm carries a negative charge of 4.2×10−7 C4.2 \times 10^{-7}\, C spread uniformly over it surface. The electric field near the mid-point of the rod, at a point on its surface is

A

−8.6×105 N C−1- 8.6 \times 10^5\, N\, C^{-1}

B

8.6×104 N C−1 8.6 \times 10^4\, N\, C^{-1}

C

−6.7×105 N C−1- 6.7 \times 10^5\, N\, C^{-1}

D

6.7×104 N C−1 6.7 \times 10^4\, N\, C^{-1}

Answer

−6.7×105 N C−1- 6.7 \times 10^5\, N\, C^{-1}

Explanation

Solution

Here, l=2.4 ml = 2.4\, m, r=4.6 mm=4.6×10−3 m r = 4.6 \,mm = 4.6 \times 10^{-3}\, m q=−4.2×10−7 Cq = -4.2 \times 10^{-7}\, C Linear charge density, λ=ql\lambda=\frac{q}{l} =−4.2×10−72.4=\frac{-4.2\times10^{-7}}{2.4} =−1.75×10−7 Cm−1=-1.75\times10^{-7}\,Cm^{-1} Electric field. E=λ2πε0rE=\frac{\lambda}{2\pi\varepsilon_{0}r} =−1.75×10−72×3.14×8.854×10−12×4.6×10−3=\frac{-1.75\times10^{-7}}{2\times3.14\times8.854\times10^{-12}\times4.6\times10^{-3}} =−6.7×105 N C−1=-6.7\times10^{5}\,N\,C^{-1}